What is the value of the infinite product: $(1+ \frac{1}{1^1}) (1+ \frac{1}{2^2}) (1+ \frac{1}{3^3}) \cdots $? What is the value of the following infinite product? 
$$\left(1+ \frac{1}{1^1}\right) \left(1+ \frac{1}{2^2}\right) \left(1+ \frac{1}{3^3}\right) \cdots  $$
Is the value known?
 A: I'm not sure what the criterium for a full answer is here, so here a technique for $(1+c_k)$ kind of products, turning the infinite product into an infinite sum: 
Via telescoping, for friendly $a_n$ and any $m$, we have
$$\lim_{n\to\infty}a_n=a_m+\sum_{n=m}^\infty\left(\dfrac{a_{n+1}}{a_n}-1\right)\,{a_n}.$$
So define
$$a_n:=\prod_{k=1}^{n-1}\left(1+c_k\right)\hspace{.5cm}\implies\hspace{.5cm}\dfrac{a_{n+1}}{a_n}-1=c_n,$$
and then
$$\prod_{n=1}^\infty\left(1+c_n\right) = \lim_{n\to\infty}a_n = \prod_{k=1}^{m-1}\left(1+c_k\right)+\sum_{n=m}^\infty c_n\prod_{k=1}^{n-1}\left(1+c_k\right).$$
For $c_n=\dfrac{1}{n^n}$, that's
$$\frac{1^1+1}{1^1}\,\frac{2^2+1}{2^2}\frac{3^3+1}{3^3}+\sum_{n=4}^\infty\frac{1}{n^n}\prod_{k=1}^{n-1}\left(1+\frac{1}{k^k}\right)=2.603\dots$$
The first term is the lower bound $\frac{70}{27}=2.592\dots$ that's been pointed out in the comment and the remaining sum $\frac{1}{4^4}\dots+\frac{1}{5^5}\dots$ collects some $\mathcal{O}(10^{-2})$.
Truncation of the product after $k=1$ reveals the infinite product is almost two times Sophomore's dream:
$$\prod_{n=1}^\infty\left(1+n^{-n}\right)\approx 2\sum_{n=1}^\infty \frac{1}{n^n}=2.582\dots$$
