Q:

Prove that $ \lim_{n \to \infty} \frac{{(\prod_{a=1}^{n} {a^{a^{p}}}})^\frac{1}{n^{p+1}}}{n^{\frac{1}{p+1}}} = e^{-\frac{1}{(p+1)^2}}$

Prove that
$$ \lim_{n \to \infty} \frac{{(\prod_{a=1}^{n} {a^{a^{p}}}})^\frac{1}{n^{p+1}}}{n^{\frac{1}{p+1}}} = e^{-\frac{1}{(p+1)^2}}$$
I tried taking ln of the limit and applying stolz cesaro but I didn't get it.

A:

If you take the ln you can rewrite it as : 
$\frac{1}{n}\sum_{a=1}^{n}(\frac{a}{n})^{p}\ln(\frac{a}{n}) + (\sum_{a=1}^{n}(\frac{a}{n})^{p}-\frac{1}{p+1})\ln(n) $
Now if you use Riemann integral property you find that the sum on the left converges to $\frac{-1}{(p+1)^2}$ and the right term converges to zero hence the result.  

