At MIT all departments have numbers, and math is 18. Last year MIT math majors produced a tee shirt that said ${i\choose 18}$ ("I choose 18") on the front, and on the back $$ \frac{34376687+1499084559i}{14485008384}. $$ With the more natural denominator $18!$ this is $$ \frac{15194495654000+662595375078000i}{18!}. $$ This suggests the question: for any $n\geq 1$ find a "nice" combinatorial interpretation of the real and imaginary parts of $i(i1)(i2)\cdots (in+1)=f_n+ig_n$. It is easy to express $f_n$ and $g_n$ as certain alternating sums of Stirling numbers of the first kind, but I don't consider this "nice." The $g_n$'s seem to alternate in sign beginning with $n=5$. The $f_n$'s alternate in sign up to $n=17$ and then seem to alternate in sign beginning with $n=18$. It is curious that $i(i1)(i2)(i3)=10$, a real number. One could ask the same question with $i$ replaced by any Gaussian integer $a+bi$. One can also ask about the asymptotic rate of growth of $f_n$ and $g_n$. Clearly $f_n^2+g_n^2\sim C\cdot (n1)!^2$, so one would expect $f_n$ and $g_n$ to be roughly of the size of $(n1)!$.

Asymptotics: Lets look at the quantity $$S(n)=(1)^{n}(n+1)\binom{i}{n+1}=i\prod_{k=1}^{n+1}\left(1\frac{i}{k}\right).$$ It's just your binomial coefficient above with the $(1)^{n+1}$ factored in, and an extra $n+1$ so it factors nicely as a product.
In particular, the angle moves around the circle like $\log n$. Application to your question: The above claim shows that $$f_{n+1} = (1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}}\cos(\log n+C_0)\left(1+O\left(\frac{1}{n}\right)\right)$$ and $$g_{n+1} \sim (1)^{n+1} n! \sqrt{\frac{\sinh{\pi}}{\pi}} \sin(\log n+C_0)\left(1+O\left(\frac{1}{n}\right)\right).$$ In particular, the ratio $g_n/f_n$ can be made arbitrarily large or small. Proof of the claim: We first note that the size is $$\sqrt{\prod_{k=1}^{n}\left(1+\frac{1}{k^{2}}\right)}=\sqrt{\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right)}+O\left(\frac{1}{n}\right).$$ To evaluate this product, recall the Weierstrass product for the Gamma function $$\left(\Gamma(z)\right)^{1}=ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k^{2}}\right)e^{\frac{z}{r}}.$$ From this it follows that $$\frac{1}{\Gamma(i)^{2}}=\frac{1}{\Gamma(i)\Gamma(i)}=\prod_{k=1}^{\infty}\left(1+\frac{1}{k^{2}}\right).$$ Using the identity $$\Gamma(x)\Gamma(x)=\frac{\pi}{x\sin\left(\pi x\right)},$$ we now have that $$\frac{1}{\Gamma(i)\Gamma(i)}=\frac{i\sin(i\pi)}{\pi}=\frac{\sinh(\pi)}{\pi},$$ which gives rise to the $\sqrt{\frac{\sinh(\pi)}{\pi}}$ term. Moving on to the evaluation of the angle, by looking at each triangle, and noting that the argument is additive when multiplied, we get that the argument equals $$\sum_{k=1}^{n}\tan^{1}\left(\frac{1}{k}\right).$$ The negative sign arises since we are working in the fourth quadrant. By looking at the Taylor series for $\tan^{1}$ we see that the above is $\log n+O(1)$, however, I would like to compute this argument more precisely, and obtain the constant. Lets compare our $\tan^{1}$ series to the harmonic series. Rewriting things in terms of a Riemann Stieltjes integral, and using summation by parts, we have that $$\sum_{k=1}^{n}\tan^{1}\left(\frac{1}{k}\right)=\int_{0}^{n}\tan^{1}\left(\frac{1}{x}\right)d\left[x\right]=[n]\tan^{1}(1/n)\int_{0}^{n}\frac{\left[x\right]}{1+x^{2}}dx. $$ Pulling out the main term with the identity $[x]=x\{x\}$, the above equals $$\int_{0}^{n}\frac{x}{1+x^{2}}dx\int_{0}^{n}\frac{\{x\}}{1+x^{2}}dx.$$ Since the first integral evaluates to $\frac{1}{2}\log(1+x^2)$, we have that $$\sum_{k=1}^{n}\tan^{1}\left(\frac{1}{k}\right)=\log n +1\int_0^\infty \frac{\{x\}}{1+x^2}dx +O\left(\frac{1}{n}\right).$$ Acknowledgements: I would like to thank Noam Elkies for pointing out that $$\prod_{k=1}^\infty \sqrt{1+\frac{1}{k^2}}=\frac{1}{\Gamma(i)}=\sqrt{\frac{\sinh(\pi)}{\pi}}$$ in the comments. Edit: Fixed the constants appearing. Interestingly $$\Gamma(i)=\sqrt{\frac{\pi}{\sinh{\pi}}}\exp\left(i\left(\frac{\pi}{2}1+\int_0^\infty \frac{\{x\}}{1+x^2}dx \right)\right).$$ 


There is no need to reinvent the wheel by estimating $\prod_{k<n}(1+\frac1{k^2})$. The asymptotic formula for $f_n + i g_n$ follows readily from Stirling's approximation (as I already noted in my comment to the original question), and indeed the same is true for the asymptotics as $n \rightarrow \infty$ of $w \choose n$ for any $w \in {\bf C}$; the answer is simply $$ \phantom{*0000000000000000000} {w \choose n} = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) \frac{(1)^n}{\Gamma(w)} n^{w1} \phantom{0000000000000000000}(*) $$ (and the $O(1/n)$ can be refined to an asymptotic series in powers of $1/n$). Note that this gives zero precisely for the values $w=0,1,2,3,\ldots$ for which $w$ is a pole of $\Gamma$, which are also the $w$ for which ${w \choose n} = 0$ for sufficiently large $n$. For $w=i$, we recover the observed behavior: $\Gamma(i)$ is a complex number of absolute value $(\pi / \sinh \pi)^{1/2}$ [in general $$\Gamma(it) = (\Gamma(it)\Gamma(it))^{1/2} = \left(\frac \pi {t \phantom. \sinh \pi t} \right)^{1/2} $$ for real $t \neq 0$], and $n^{w1}$ is a complex number of absolute value $1/n$ that goes once around the origin when $n$ increases by a factor $e^{2\pi}$. Thus each of $\lbrace f_n \rbrace$ and $\lbrace g_n \rbrace$ alternates in sign outside an infinite sequence of exceptions that's asymptotically a geometric sequence with common ratio $e^\pi$. To prove $(*)$, write $$ {w \choose n} = \frac{(1)^n}{n!} \prod_{k=0}^{n1} (kw) = \frac{(1)^n}{n!} \frac{\Gamma(nw)}{\Gamma(w)} = \frac{(1)^n}{n\Gamma(w)} \frac{\Gamma(nw)}{\Gamma(n)}. $$ Now we understand $(1)^n/n$, and the factor $1 / \Gamma(w)$ is constant, so we're left with $\Gamma(nw) / \Gamma(n)$. We apply the following form of Stirling's formula: there exists a constant $\varpi>0$ (known to equal $2\pi$, but we shall not need this) such that $$ \Gamma(z) = \bigl(1 + O(z^{1}\bigr) z^z e^{z} \sqrt{\varpi/z} $$ holds as $z \rightarrow \infty$ in the right halfplane, where $z^z = \exp (z \log z)$ and $\sqrt{\varpi/z}$ are defined using the principal branches of $\log z$ and $\sqrt z$. This gives $$ \frac{\Gamma(nw)}{\Gamma(n)} = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) \frac{(nw)^{nw} e^{(nw)} (\varpi/(nw))^{1/2}} {n^n e^{n} (\varpi/n)^{1/2}} $$ $$ = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) \frac{(nw)^{nw}}{n^n} e^w \left(1\frac{w}{n}\right)^{1/2}. $$ Now the last factor is $1 + O(1/n)$; the factor $e^w$ is constant; and $$ (nw)^{nw} = (nw)^{w} (nw)^n = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{w} (nw)^n. $$ So we're left with $$ \frac{\Gamma(nw)}{\Gamma(n)} = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{w} e^{w} \left(1  \frac{w}{n}\right)^n = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{w}. $$ This completes the proof of $(*)$ (and the cancellation in the last step leads me to suspect that even this use of Stirling is more complicated than necessary). 

