Jokes in the sense of Littlewood: examples? First, let me make it clear that I do not mean jokes of the
"abelian grape" variety. I take my cue from the following 
passage in A Mathematician's Miscellany by J.E. Littlewood
(Methuen 1953, p. 79):

I remembered the Euler formula $\sum n^{-s}=\prod (1-p^{-s})^{-1}$;
  it was introduced to us at school, as a joke (rightly enough, and
  in excellent taste).

Without trying to define Littlewood's concept of joke, I venture
to guess that another in the same category is the formula

$1+2+3+4+\cdots=-1/12$,

which Ramanujan sent to Hardy in 1913, admitting "If I tell you this 
you will at once point out to me the lunatic asylum as my goal."
Moving beyond zeta function jokes, I would suggest that the empty
set in ZF set theory is another joke in excellent taste. Not only
does ZF take the empty set seriously, it uses it to build the whole
universe of sets.
Is there an interesting concept lurking here -- a class of
mathematical ideas that look like jokes to the outsider, but which
turn out to be important? If so, let me know which ones appeal to
you.
 A: If $1-ab$ is invertible for $a$, $b$ in a (noncommutative) ring then so is $1-ba$.
Proof:
$$(1-ba)^{-1} = 1+ba +baba+\cdots = 1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a,$$
The meaningless infinite series  give the right answer (which is hard to guess).
A: The Cayley-Hamilton Theorem:
If $A$ is a square matrix with characteristic polynomial $p(\lambda) = \det(A-\lambda I)$, then $p(A) = 0$.
Because you know, you "just plug in."
A: The chain rule "joke" reminded me of a similar notation joke: Radon-Nikodym derivatives.
If $\mu$, $\nu$, $\lambda$ are $\sigma$-finite measures with $\nu \ll \mu \ll \lambda$, and $f \geq 0$ is measurable, then:
$$\int f\ d\nu = \int f \left[\frac{d\nu}{d\mu}\right]\ d\mu$$
and
$$\left[\frac{d\nu}{d\lambda}\right] = \left[\frac{d\nu}{d\mu}\right]\left[\frac{d\mu}{d\lambda}\right]$$
A: In the same vein as the "Freshman's dream"
$$(a + b)^p = a^p + b^p,$$
which is true in characteristic $p$, there is also the "Sophomore's dream", which is the identity
$$\int_{0}^{1}{x^{-x} \: dx} = \sum_{n = 1}^{\infty}{n^{-n}}.$$
Surprisingly enough, this identity is actually correct.
A: Hausdorff dimension.  Try showing a Sierpinski triangle to a non-mathematician and explaining that it is a 1.585-dimensional object.
A: $$\dfrac{16}{64} = \dfrac{1\not{6} }{\not{6} 4} = \dfrac{1}{4}$$
$$25^{1/2} = \not25^{1/\not2} = 5^1 = 5$$
$$\sqrt[6]{64} = \sqrt[\not 6]{\not 64} = \sqrt{4}$$
A: In german elementary schools you learn "In Summen kürzen nur die Dummen.", i.e. it is dumb to cancel sums in quotients. But later in algebra or category theory you may learn
$\sum_{i \in I} A_i / \sum_ {i \in I} B_i = \sum_{i \in I} A_i / B_i$,
which is correct.
A: In Cardano's formula for the roots of the cubic there are negative numbers under the square roots if and only if all roots of the polynomial are real.  That, and the fact that Cardano and his contemporaries didn't even believe in negative numbers.
A: Tim Gowers mentioned infinities that may sound like jokes, especially to outsiders. Here is one specific example: you are standing in a room; at every tick of the clock, someone throws in a pair of numbered ping-pong balls: 1 & 2, then 3 & 4, etc... and you only have enough time to throw out one of them before the next tick. If you throw out the one with the largest number, then after $\omega$ ticks of the clock, you are in the room with all the odd-numbered balls, whereas if you always threw out the ball with the smallest number, you would be rid of them all!
And what if the balls are not numbered? A good way to get non-mathematicians thinking about infinity.
A: The fundamental axioms of mathematics are inconsistent if
and only if we can prove that they are consistent.
(Because, you know, it follows from "logic." See Second
Incompleteness
theorem)
A: Nobody mentioned the third isomorphism theorem yet? If $B$ and $C$ are normal subgroups of $A$ and $C \le B$ then $\frac{A/C}{B/C} \cong \frac{A}{B}$.
A: The yoga of motives, in the parts which are conjectural but inspire the right guesses, is a huge joke!
A: Gerry's answer reminded me of the "Lucky Larry" series, which is a regular column in The AMATYC Review.
One particularly nice one is
$$\lim_{x\to\infty}\frac{\ln\ln x}{\ln x}$$
where you can cancel the $\ln x$ and wind up with $\ln 1=0$, which turns out correct.
A: Many infinite, periodic expression involving numbers is a joke, according to non-mathematicians. Three examples:
$$0,999999999\cdots=1,$$
$$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}=\frac{1+\sqrt5}{2},$$
$$\sqrt{-a^2+\sqrt{-a^2+\sqrt{-a^2+\cdots}}}=\frac12(1+\sqrt{1-4a^2}).$$
A: Two ways to calculate $1+2+3+\dots$
1st way. Let $$1-1+1-1+\dots=a.$$ Then also $$0+1-1+1-1+\dots=a.$$ Sum up (summing respective terms), get $$2a=1+0+0+\dots=1,$$ $a=1/2$.
Let now $$1-2+3-4+\dots=b.$$ Again, write down $$0+1-2+3-\dots=b.$$ Sum up and get  $$2b=1-1+1-1+\dots=1/2,$$ 
$b=1/4$. Now let 
$$
1+ 2+3+ 4+5+\dots=с.
$$
Then 
$$
0+4\cdot 1+0+4\cdot 2+0+4\cdot 3+0+\dots=4с.
$$
Substract and get $$-3с=1-2+3-4+5+\dots=1/4,$$ hence
$$1+2+3+\dots=-1/12.$$
2nd way. Let  $$1+2+3+\dots=S.$$ Then 
$$
2+4+6+\dots=2\cdot(1+2+3+\dots)=2S.
$$ Hence
$$
1+3+5+\dots=(1+2+3+4+5+\dots)-
(0+2+0+4+0+\dots)=-S.
$$
Now sum up $$1+2+3+4+\dots=S$$ and $$0+1+3+5+\dots=-S.$$ We get 
$$
0=1+3+6+9+\dots=1+3\cdot(1+2+3+\dots)=1+3S,
$$ 
$S=-1/3$. So
$$
1+2+3+\dots=-1/3.
$$
The question is: why the first way gives correct answer $\zeta(-1)=-1/12$, while the second leads to incorrect answer?
A: Another example from intro calculus: I once put a question of the form "$y=f(x)^{g(x)}$, find $y'$" on an exam. One student reasoned, if the exponent were a constant, the answer would be $g(x)f(x)^{g(x)-1}f'(x)$, but that's not right; if the base were a constant, the answer would be $g'(x)f(x)^{g(x)}\log f(x)$, but that's not right either; so I'll put them together to get $$g(x)f(x)^{g(x)-1}f'(x)+g'(x)f(x)^{g(x)}\log f(x)$$This joke was on me, since that turns out to be correct.
A: Mazur's proof that knots do not have inverses under addition of knots:
If $A+B=0$, then $$A = A + (B+A)+(B+A)+\cdots=(A+B)+(A+B)+\cdots=0.$$
This is like the traditional joke proof that $1=0$ with $A=1$, $B=-1$; the difference is that the proof with knots is valid because the infinite sums of knots are meaningful: make the knots smaller and smaller. 
A: The field with one element seems a good example.
A: The chain rule, in the form $${dy\over dx}={dy\over du}{du\over dx}$$ is a joke - you just cancel the $du$, top and bottom. 
A: I've always thought of compactness arguments for passing from a finite result to an infinite one as sorts of jokes. The general idea, after all, is that to get the infinite analogue of the finite result, you just "make it bigger and bigger!" (Of course it doesn't work in all situations, but when it does it's often forehead-slappingly simple.)
A: A typical proof of Kolmogorov's zero-one law has as its punchline 'therefore A is independent of A'.  Perhaps not a joke in the sense of Littlewood, but amusing nonetheless.
A: I want to evaluate $f(x+t)$. This is a function of two variables, but let's consider it a function $F(t)$ whose value is a function of $x$, i. e., $F(t)(x) = f(x+t)$. Note that $F(0) = f$, and in general $F$ satisfies the differential equation 
$$F'(t) = D_x(F(t))$$ (both sides being the function $x\mapsto f'(x+t)$). But $D_x$ is just a linear operator, so this is just a homogeneous linear ODE with constant coefficients. The solution is thus $$f(x+t) = F(t)(x) = (e^{tD_x}F(0))(x) = (e^{tD_x}f)(x)
= \sum_{n=0}^\infty \frac{((tD_x)^nf)(x)}{n!}
= \sum_{n=0}^\infty \frac{t^n f^{(n)}(x)}{n!}.$$ Voilà, Taylor series! 
A: A good joke about infinity is the following. A hotel has rooms $1,2,\dots$. Every room is full when a new guest arrives. The clerk moves the occupant of room $n$ to $n+1$ to make room for the new guest in room 1. An hour later another guest arrives and the clerk repeats the 
process. 30 minutes later a third guest arrives and the process is repeated. Then 15 minutes, 7.5 minutes, etc., until two hours after the first new guest infinitely many guests have arrived and been accommodated. The clerk is very pleased with himself for dealing with these infinitely many guests, when he notices to his horror that all the rooms are empty! All the guests have mysteriously disappeared!
A: An expansion on Timothy Chow's example of Grandi's series $1 - 1 + 1 - 1 \pm \cdots = \frac{1}{2}$.  It is possible to interpret the left hand side as computing the Euler characteristic of infinite real projective space $\mathbb{R}P^{\infty}$, which is a $K(\mathbb{Z}/2\mathbb{Z}, 1)$ and therefore rightfully has orbifold Euler characteristic $\frac{1}{2}$!  I think I learned this example from somewhere on Wikipedia.
A: We owe Paul Dirac two excellent mathematical jokes. I have amended them with a few lesser known variations.
A. Square root of the Laplacian: we want $\Delta$ to be $D^2$ for some first order differential operator (for example, because it is easier to solve first order partial differential equations than second order PDEs). Writing it out,
$$\sum_{k=1}^n \frac{\partial^2}{\partial x_k^2}=\left(\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\right)\left(\sum_{j=1}^n \gamma_j \frac{\partial}{\partial x_j}\right) = \sum_{i,j}\gamma_i\gamma_j \frac{\partial^2}{\partial x_i x_j},
$$
and equating the coefficients, we get that this is indeed true if
$$D=\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\quad\text{and}\quad \gamma_i\gamma_j+\gamma_j\gamma_i=2\delta_{ij}.$$ (With a $2$ on the right hand side.)
It remains to come up with the right $\gamma_i$'s. Dirac realized how to accomplish it with $4\times 4$ matrices when $n=4$; but a neat follow-up joke is to simply define them to be the elements $\gamma_1,\ldots,\gamma_n$ of
$$\mathbb{R}\langle\gamma_1,\ldots,\gamma_n\rangle/(\gamma_i\gamma_j+\gamma_j\gamma_i - 2\delta_{ij}).$$

Using symmetry considerations, it is easy to conclude that the commutator of the $n$-dimensional Laplace operator $\Delta$ and the multiplication by $r^2=x_1^2+\cdots+x_n^2$ is equal to $aE+b$, where
$$E=x_1\frac{\partial}{\partial x_1}+\cdots+x_n\frac{\partial}{\partial x_n}$$ is the Euler vector field. A boring way to confirm this and to determine the coefficients $a$ and $b$ is to expand $[\Delta,r^2]$ and simplify using the commutation relations between $x$'s and $\partial$'s. A more exciting way is to act on $x_1^\lambda$, where $\lambda$ is a formal variable:
$$[\Delta,r^2]x_1^{\lambda}=((\lambda+2)(\lambda+1)+2(n-1)-\lambda(\lambda-1))x_1^{\lambda}=(4\lambda+2n)x_1^{\lambda}.$$
Since $x_1^{\lambda}$ is an eigenvector of the Euler operator $E$ with eigenvalue $\lambda$, we conclude that
$$[\Delta,r^2]=4E+2n.$$

B. Dirac delta function: if we can write
$$g(x)=\int g(y)\delta(x-y)dy$$
then instead of solving an inhomogeneous linear differential equation $Lf=g$ for each $g$, we can solve the equations $Lf=\delta(x-y)$ for each real $y$, where a linear differential operator $L$ acts on the variable $x,$ and combine the answers with different $y$ weighted by $g(y)$. Clearly, there are fewer real numbers than functions, and if $L$ has constant coefficients, using translation invariance the set of right hand sides is further reduced to just one, $\delta(x)$. In this form, the joke goes back to Laplace and Poisson.

What happens if instead of the ordinary geometric series we consider a doubly infinite one? Since
$$z(\cdots + z^{-n-1} + z^{-n} + \cdots + 1 + \cdots + z^n + \cdots)= 
\cdots + z^{-n} + z^{-n+1} + \cdots + z + \cdots + z^{n+1} + \cdots,$$
the expression in the parenthesis is annihilated by the multiplication by $z-1$, hence it is equal to $\delta(z-1)$. Homogenizing, we get
$$\sum_{n\in\mathbb{Z}}\left(\frac{z}{w}\right)^n=\delta(z-w)$$
This identity plays an important role in conformal field theory and the theory of vertex operator algebras.

Pushing infinite geometric series in a different direction,
$$\cdots + z^{-n-1} + z^{-n} + \cdots + 1=-\frac{z}{1-z} \quad\text{and}
\quad 1 + z + \cdots + z^n + \cdots = \frac{1}{1-z},$$
which add up to $1$. This time, the sum of doubly infinite geometric series is zero!
Thus the point $0\in\mathbb{Z}$ is the sum of all lattice points on the non-negative half-line and all points on the positive half-line:
$$0=[\ldots,-2,-1,0] + [0,1,2,\ldots] $$
A vast generalization is given by Brion's formula for the generating function for the lattice points in a convex lattice polytope $\Delta\subset\mathbb{R}^N$ with vertices $v\in{\mathbb{Z}}^N$ and closed inner vertex cones $C_v\subset\mathbb{R}^N$:
$$\sum_{P\in \Delta\cap{\mathbb{Z}}^N} z^P = 
\sum_v\left(\sum_{Q\in C_v\cap{\mathbb{Z}}^N} z^Q\right),$$
where the inner sums in the right hand side need to be interpreted as rational functions in $z_1,\ldots,z_N$.

Another great joke based on infinite series is the Eilenberg swindle, but I am too exhausted by fighting the math preview to do it justice.
A: Multiplication is repeated addition:
$$
 x^2 = \underbrace{x+\cdots+x}_{x\text{ times}}
$$
Differentiate (remember the chain rule for partial derivatives):
$$\frac{d}{dx} \big(x^2\big)\qquad\qquad$$ 
$$ \qquad= {\underbrace{1+\cdots+1}_{x\text{ times}}}$$
$$\qquad{}+ \underbrace{x+\cdots+x}_{1\text{ times}}$$
$$ \qquad= x + x = 2x .$$
A: In characteristic $p$, the so-called biologists' rule
$$(a+b)^p = a^p + b^p$$
 (which got its name by mathematics students that worked as teaching assistants for "mathematics for biologists") is correct.
A: There are other divergent series that fit the bill, such as $1-1+1-1+ \cdots = 1/2$.  Here's one from formal language theory: Suppose we define a language $L$ recursively by the rule $L = 1 | aL$, meaning that the empty string $1$ is in $L$, and the letter $a$ followed by any element in $L$ is also in $L$.  Jokingly, we note that $|$ is akin to addition and concatenation is akin to multiplication, so we can solve for $L$: $1 = L - aL  = L (1-a)$, so
$$L = {1\over 1-a} = 1|a|aa|aaa|aaaa \ldots,$$
which is the right answer.
The umbral calculus could also be considered an elaborate joke.
A: The classical Stokes formula $\int_{\partial\Omega}\omega=\int_\Omega d\omega$ is certainly a Littlewood type joke. That is especially true if you learn it after you've spent a few months covering vector calculus, learned rotor, divergence, path and surface integrals of 2 kinds, etc., which is the standard route to follow.
A: In a probability oral exam, a student is asked to compute the probability that a random number 
chosen from the interval $[0,1]$ is larger than $2/3$. The students answers $1/3$. The teacher asks him to explain his argument, and he says: well, there are three possibilities: the number is either less than, or bigger than, or equal to $2/3$, so, the probability is $1/3$!
A: How many finite sets are there?
Well, there is one set of cardinality 0; one set of cardinality one; one set of cardinality 2, but since its automorphism group has order 2, we only count it with multiplicity 1/2; there is one set of cardinality 3, counted with multiplicity 1/3!; ...
So the number of sets is
$$
1 + 1+ 1/2! + 1/3! + \dots = e
$$
A: This is a souped-up version of the freshman's dream: as Jon Borwein pointed out to me: if $a_n=(-1)^n/(2n+1)$, then $$\left(\sum_{n=-\infty}^{\infty}a_n\right)^2=\sum_{n=-\infty}^{\infty}a_n^2$$ as they are both $\pi^2/4$. 
Moreover, this can be proved by telescoping sums:
$$
\begin{align}
\left( \sum_{n} \frac{(-1)^n}{2n+1} \right)^2 &= \sum_{m,n} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \\
&= \sum_{m,k} \frac{(-1)^k}{(2m+1)(2m+2k+1)} \\
&= \sum_{m} \frac{1}{(2m+1)^2} + \sum_{k \neq 0 } \sum_m \left( \frac{(-1)^k}{2 k (2m+1)} - \frac{(-1)^k}{2 k (2m+2k+1)} \right) \\
&= \sum_{m} \frac{1}{(2m+1)^2}
\end{align}$$
Of course, one needs to justify rearranging the conditionally convergent series, but that spoils the joke.
A: The Eilenberg swindle was briefly mentioned by Victor Protsak, and something which always struck me as similar (and seemed like black magic to me when I first saw it) is the Pelczynski decomposition method for proving that complemented subspaces of $\ell^p$ are isomorphic to $\ell^p$. I don't have a link to hand but might try to write something out later (or someone else is welcome to edit this).
A: When solving the linear recursive equation
$a_{n+2} = a_{n+1} + 2 a_n$, you solve the quadratic equation 
$x^2 = x + 2$, which has the solutions $x= 2$ and $x = -1$, and then you get a basis of solutions $2^n$ and $(-1)^n$.
What if you start with $a_{n+2} = 4a_{n+1} - 4a_n$, so that the quadratic equation
$x^2 = 4x - 4$ has only one solution $x = 2$? Easy, you take the solution $2^n$, and its derivative with respect to 2, i.e. $n \cdot 2^{n-1}$.
A: You can't prove something exists just by computing the probability of its existence - right? The first application of the  "probabilistic arguments" in Combinatorics I have encountered was  this; took me a long time to get it.  
A: (An example from set theory, more specifically: forcing): 
The backslash $\backslash$ is used for set difference, the forward slash $/$ is often used for taking a quotient.   In my experience, students often have a hard time distinguishing between the two.  (Between the two symbols, not the two concepts.) 
But there is one case where the difference does not matter:  if $B$ is a Boolean algebra, and $I$ an ideal, then the sets $B\setminus I$ (= $I$-positive elements)  and $B/I$ (quotient Boolean algebra) are equivalent as forcing notions:  $B\setminus I  = B/I $. 
For example, random forcing can be seen as "Borel sets modulo measure zero sets", or equivalently as "Lebesgue-positive Borel sets (without any identification)". 
A: I know you say "moving beyond zeta function jokes", but I'd say the following two zeta-regularizations deserve to be alongside your Ramanujan example:
$$\infty!= \sqrt{2\pi}\qquad\qquad\mbox{and }\qquad\qquad
\prod_{\mbox{$p$ prime}}p =4\pi^2.$$
One can also entertain beginning calculus students with $\frac{1}{2}!=\frac{1}{2}\sqrt{\pi}$ as a way of introducing the Gamma function.
A: Given a function $f$ on the real line, let's compute the function
$\sum f$, taking $n \mapsto f(1) +\cdots + f(n)$. Well, $\sum = 1/\Delta$,
where $\Delta$ is the differencing operator $\text{shift} - 1$. And the shift
operator is the exponential of the differentiation operator (this being,
essentially, Taylor's theorem). Hence
$$ \sum = \frac{1}{e^D - 1} = \frac{1}{D} \frac{D}{e^D-1} $$
Using L'Hopital's rule on the latter as $D\to 0$, 
whatever THAT means, we see the limit is $1$. So expand in a power series:
$$ \frac{1}{D} \frac{D}{e^D-1} = \frac{1}{D} (1 + \text{power series in $D$}) $$
The first term is $1/D$, which is of course $\int$. 
No surprise:
$\sum = \int + $ correction terms. What the above suggests is that those 
correction terms come from the Taylor expansion of $\frac{D}{e^D - 1}$.
This leads to the Euler summation formula (and eventually, to
Hirzebruch-Riemann-Roch). 
I learned this from "Concrete Mathematics", where I recall this joke
being attributed to Laguerre. Part of why it is a joke is that the Euler
summation formula has an error term, that can't be neglected for most
functions, e.g. $\ln(x)$ which one wants to sum up to compute $\ln(n!)$. 
It can be neglected for polynomials times exponentials.
A: Let $C(x) = \sum_{n \ge 0} \frac{1}{n+1} {2n \choose n} x^n$ be the generating function for the Catalan numbers.  Then $C(x) = \frac{1 - \sqrt{1 - 4x}}{2}$.  In particular, $C(1) = \frac{1 - \sqrt{-3}}{2}$ is a sixth root of unity - except of course that $C(1)$ does not converge.
Nevertheless, there is a remarkable sense in which $C^7 = C$; see Blass' Seven Trees in One and the generalization in Fiore and Leinster's Objects of Categories as Complex Numbers.  As with many results of this type, I learned about this from John Baez's TWF 202.
A: Lagrange's equation:
$$\frac{d}{dt} \left( \frac{\partial}{\partial \dot{q}} \mathcal{L} \right) =\frac{\partial}{\partial q} \mathcal{L},$$
as $\dot{q}=\frac{dq}{dt}$, you can simplify "$dt$".
A: I do a double take every couple of months when I remember that
$(1 + 2 + 3 + \ldots + n)^2 = 1^3 + 2^3 + 3^3 + \ldots + n^3$
It just seems plain weird.
A: I can't say I share Littlewood's sense of humor, but here is a formula that made me grin the first time I saw it ($A$ and $B$ are non-commuting square matrices):
\begin{equation}\frac{d}{dt}e^{A+B t}=\int_0^1 e^{s(A+B t)}\ B\ e^{(1-s)(A+B t)} \ ds.
\end{equation}
A: I'm not sure if this counts, but here's one of my own devising:
How do we find second derivatives of inverse functions?
Well, the second derivative of $y = f(x)$ is defined as $(f')' = \frac{d\frac{dy}{dx}}{dx}$, and by the quotient rule we can write this as $\frac{dxd^2y-dyd^2x}{dx^3} = \frac{d^2y}{dx^2}-\frac{dyd^2x}{dx^3}.$ (Since $x$ usually varies linearly, we normally substitute $d^2x = 0$.)
So by symmetry, the second derivative of the inverse function is $(f^{-1})'' = \frac{d^2x}{dy^2}-\frac{dxd^2y}{dy^3} = -\frac{dx^3}{dy^3}(\frac{d^2y}{dx^2}-\frac{dyd^2x}{dx^3}) = -f''/(f')^3.$
We can also directly derive the chain rule for second derivatives: Let $x = f(v)$, $v = g(u)$, and we get
$(f\circ g)'' = \frac{d^2x}{du^2} - \frac{dxd^2u}{du^3} = (\frac{d^2x}{dv^2}-\frac{dxd^2v}{dv^3})\frac{dv^2}{du^2}+\frac{dx}{dv}(\frac{d^2v}{du^2}-\frac{dvd^2u}{du^3}) = (f''\circ g)(g')^2 + (f'\circ g)g''.$

Here's another one:
A friend of mine (Sam Elder) was trying to calculate the number of combinations on a simplex lock with $n$ buttons. After some work he had gotten a recurrence which I'm going to write as $2A_m = 1+\sum_{k=0}^m \binom{m}{k}A_{k}$, which he showed to me. My thought process went like this:
Hmm, this looks like the recurrence for the Bernoulli numbers. How did we prove the recurrence for the Bernoulli numbers again? One way is to use the well-known fact that $\sum_{n=0}^{\infty} n^m = -\frac{B_{m+1}}{m+1}$, so
$-\frac{B_{m+1}}{m+1} = \sum_{n=0}^{\infty} n^m = \sum_{n=0}^{\infty}(n+1)^m = -\sum_{k=0}^m\binom{m}{k}\frac{B_{k+1}}{k+1}$.
Working backwards, what I need to do is find a sequence of functions $f_m(n)$ satisfying $2f_m(n+1) = \sum_{k=0}^m\binom{m}{k}f_{k}(n)$. This leads naturally to the choice $f_m(n) = \frac{n^m}{2^n}$. So the number of simplex combinations on m buttons is
$A_m = \sum_{n=1}^{\infty} \frac{n^m}{2^n}.$
Sam did not find this formula incredibly helpful.
(Continuing the analogy with the Bernoulli numbers, we can also derive the formula
$\sum_{n=1}^x \frac{n^m}{2^n} = A^m-\frac{(x+A)^m}{2^x},$
where we interpret $A^m$ as $A_m$.)
A: For consecutive Farey fractions $\frac{a}{b}, \frac{c}{d}$ the mediant is obtained via a "simple's man addition":
$$ \frac{p}{q} = \frac{a+c}{b+d} $$
which since $\frac{a}{b},\frac{c}{d}$ are consecutive if and only if $det\begin{pmatrix}a & c\\\\ b & d\end{pmatrix} = 1$ also turns out to be the rule of invariance of the determinant when you add a column to another column.
A: Here's one of my own.  Maybe I should publish it?  A confused calculus student attempted to evaluate
$$
\frac{d}{dx} \left( 1^n + 2^n + 3^n +  \cdots + (x-1)^n \right)
$$
and got
$$
n1^{n-1} + n2^{n-1} + n3^{n-1} + \cdots + n(x-1)^{n-1} + \text{constant}.
$$
That is in fact correct.  The "constant" is a Bernoulli number.
A: Let "$\int$" denote $\int_0^x$. We want to find the solution to
$$\int f = f-1.$$
We simply "factor out" $f$, getting $1=\left(1-\int\right)f$. Thus, $f=(1-\int)^{-1}1$. 
Using the geometric series,
$$f=\left(1+\int+\iint+\iiint+\cdots\right)1=1+\int_0^x1~dx+\int_0^x\int_0^x1~dx+\cdots$$
Thus,
$$f=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots=e^x,$$
as expected. (This was told to me by Steve Miller)
A: This isn't a particularly interesting example, but the existence of different sizes of infinity fits your criterion of being something that makes outsiders laugh (as I know from experience) and that is also very important to mathematicians. 
The familiar argument that says that if you want an explicit example of $a^b=c$ with a and b irrational and c rational, then one of $a=b=\sqrt{2}$ or $a=\sqrt{2}^{\sqrt{2}}$ and $b=\sqrt{2}$ will work is certainly an argument that makes people laugh. Though the result itself is not very important, the phenomenon it illustrates is quite important.
Added two minutes later: I've just had a look at Scott Aaronson's post and seen that Erik, one of the earlier commenters, chose precisely the same two examples. It was a coincidence -- honest.
A: This was once presented to me as a kind of proof, though I think it works better as a kind of joke:
To compute ${\partial^n\over\partial x^n}(fg)$, split ${\partial\over\partial x}$ into the sum of a piece $D$ that just acts on $f$ (acting as the identitiy on $g$) and a piece $E$ that just acts on $g$ (acting as the identity on $f$) and write
${\partial^n\over\partial x^n}(fg) = (D+E)^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i E^{n-i}(fg) = \sum_{i=0}^n \binom{n}{i} {\partial^i f\over\partial x^i}{\partial^{n-i} g\over\partial x^{n-i}}$.
A: The geometric series expansion of projective space:  $\frac{\mathbf{C}^{n+1} - \mathrm{pt}}{\mathbf{C} - \mathrm{pt}} = \mathbf{C}^n + \cdots + \mathbf{C}^1 + \mathrm{pt}$
A: Another geometric series joke:
Let $S_\epsilon$ be the $\epsilon$-shift operator and let $D = (1 - S_\epsilon)/\epsilon$ be the formal derivative operator. Then $D^{-1} = \epsilon \sum_{n=0}^\infty {S_\epsilon}^n$, so $(D^{-1}f)(x) = \epsilon(f(x) + f(x-\epsilon) + f(x-2\epsilon) + \cdots)$, the integral of $f$ from $-\infty$ to $x$. Thus differentiation and integration are inverses.
Playing a prank with the binomial formula on $D$ yields fractional derivatives, another famous joke.
John kicked off the thread with zeta functions; my favorite zeta function joke has to be Euler's solution to the Basel problem where he factored $sin(x)$ as a power series according to its roots $\pm n\pi$.
A: The number of finite sets is $e$. 
(Since of course we should count in the stack sense: up to isomorphism, dividing by the number of automorphisms).
A: I've always seen the following identity for the determinant of block matrices as a 'joke'
$\det\left( \begin{array}{cc} A & B \newline C & D \end{array} \right) = \det(AD-BC)$
A: I suppose this is a silly example, but as far back as grade school I found it amusing (and maybe a little profound) that to invert a fraction $\frac{a}{b}$, you literally invert it! As I learned about more and more mathematical objects through the years, I kept waiting for this kind of thing to happen again, but it never really did.
A: I recall that the following simple "proof" of $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ is attributed to Euler:
Begin with the fact that for a polynomial $a_0 + a_1 x + \cdots + a_N x^N$ the sum of the inverses of the roots is given by $\sum_{n=1}^N \frac{1}{x_n} = -\frac{a_1}{a_0}$. (If you only remember the formula for the sum of the roots just make a change of variable $y=1/x$). Now consider the "polynomial" $$\frac{\sin\sqrt{x}}{\sqrt{x}} = 1 - \frac{x}{3!} + \cdots$$ whose roots are $x_n = (n\pi)^2$ for $n\in N$. By applying the aforementioned fact the desired result is immediate.
A: Another joke in the spirit of the chain rule;  you solve separable differential equations by "multiplying by g(y)dx"
$$g(y) dx \left(\frac{dy}{dx} = \frac{f(x)}{g(y)}\right) \Rightarrow g(y)dy = f(x) dx$$
Then, there's nothing to be done but integrating to get rid of the dx and dy.
I also like to point out to students who ask about cancellation in the chain rule that you can cancel there just like you can cancel the sixes and nines respectively in 
$$\frac {16}{64} = \frac 1 4 \qquad \text{and} \qquad \frac{19}{95}=\frac 1 5;$$
that is, carefully, and when it makes sense to do so.
A: Yet another divergence joke:
$$\sum_{j=0}^\infty (-1)^j j!=e E_1(1)\approx 0.596347362\dots$$
where $E_n(z)$ is an exponential integral.
This is intimately related to the formal hypergeometric series ${}_2 F_0$ being the asymptotic expansion of a certain convergent improper integral.
This also serves as a warning of sorts to users of convergence acceleration methods like Wynn ε or Levin t ; attempts to sum a divergent series might give "correct" yet still unexpected answers.
A: Here's more divergence craziness:
$$\int_0^\infty \sin\;u\mathrm{d}u=1$$
and
$$\int_0^\infty \ln\;u\;\sin\;u\mathrm{d}u=-\gamma$$
($\gamma$ is of course the Euler-Mascheroni constant)
which only makes sense when interpreted as
$$\lim_{\varepsilon\to 0} \int_0^\infty \exp(-\varepsilon u)\ln\;u\;\sin\;u\mathrm{d}u$$
and similarly for the first one.
Results obtained from numerical quadrature methods specially designed for infinite oscillatory integrals (e.g. the Ooura-Mori double exponential quadrature and the Longman scheme) agree with these closed forms.
A: If $\frac{8}{0}= \infty$, then $\frac{n}{0}= n^{'}$, where $n^{'}$ is $n$ rotated by 90 degrees on the right.
A: The journal of unpublishable mathematics, which seems to be down at the moment is one of my favourites
