How can I evalute this product? How can I evalute this product??
$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}}$$
Unfortunately, I have no idea.
 A: Hint:
Taking the logarithm,
$$\log p(n)=-\log n\sum_{i=1}^\infty i\,n^{-i}.$$
This summation, which is a modified geometric series, has a closed-form formula.
A: \begin{eqnarray*}
P=\prod_{i=1}^{\infty} (n^{-i})^{n^{-i}} = \prod_{i=1}^{\infty} n^{-in^{-i}} = n^{-\sum_{i=1}^{\infty}in^{-i}}  
\end{eqnarray*}
$\sum_{i=1}^{\infty}ix^{i}= \frac{x}{(1-x)^2}$
\begin{eqnarray*}
P=n^{-\frac{n}{(n-1)^2}}  
\end{eqnarray*}
A: HINT:
$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}} = \prod_{i=1}^{\infty} \frac{1}{n^{\frac{i}{n^i}}} = \frac{1}{n^{\frac{1}{n}}} \cdot \frac{1}{n^{\frac{2}{n^2}}}\cdot\frac{1}{n^{\frac{3}{n^3}}}...$$
$$\prod_{i=1}^{k} {(n^{-i})}^{n^{-i}} = \frac{1}{n^{\frac{n^{k-1} + 2 \cdot n^{k-2}...(k-1) \cdot n + k}{n^k}}}$$
And the exponent can be made into a summation.
A: Taking log
$$\ln\Big( \prod_{i=1}^{\infty} (n^{-i})^{n^{-i}}\Big)$$$$$$
$$\sum_{i=1}^{\infty} \frac{i}{n^i} \ln n$$$$$$
$$\ln n \sum_{i=1}^{\infty} \frac{i}{n^i} $$$$$$

Let$$$$$$S= \sum_{i=1}^{k} \frac{i}{n^i}$$ $$$$
  $$S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\cdots+\frac{k}{n^k}$$$$$$
  Multiplying by $\frac{1}{n}$$$$$
  $$\frac{S}{n}=\frac{1}{n^2}+\frac{2}{n^3}+\frac{3}{n^4}+\cdots+\frac{k}{n^{k+1}}$$$$$$
  Now $S-\frac{S}{n}$$$$$
  $$\frac{n-1}{n} S =\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\cdots+\frac{1}{n^k}-\frac{k}{n^{k+1}}$$$$$$
  Using GP formula,
  $$\frac{n-1}{n} S =\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}$$$$$$
  $$S=\frac{n}{n-1} \Biggr(\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}\Biggr)$$$$$$
  For $k=\infty$ if $n\gt 1$$$$$
  $$S=\frac{n}{n-1}\frac{1}{n} \Big(\frac{n}{n-1}\Big)$$$$$$
  $$S=n\Big(\frac{1}{n-1}\Big)^2$$$$$$

Then substitute it back

$$n \ln n  \Big(\frac{1}{n-1}\Big)^2$$Taking antilog,$$n^{n\Big(\frac{1}{n-1}\Big)^2}$$

This is only for $n\gt 1$
