<s>[INST] 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.
 [/INST] 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 $n\gt 1$
</s>