Prove $\lim\limits_{n\to\infty}n^{-1/(p+1)}(1^{1^p}\cdot 2^{2^p}\cdots n^{n^p})^{n^{-p-1}}=e^{-1/(p+1)^2}$ Prove that for real number $p\geq  0$ we have 
$$\lim_{n\to\infty}\frac{(1^{1^p} \cdot2^{2^p}\cdots n^{n^p})^\frac{1}{n^{p+1}}}{n^\frac{1}{(p+1)}} = \exp\left(\dfrac{-1}{(p+1)^2}\right).$$
 A: As several answers/hints have pointed out: consider the natural log of the LHS. We get
$$\frac{1^p\ln(1)+2^p\ln(2)+..+n^p\ln(n)}{n^{p+1}}-\frac{\ln(n)}{p+1}=\frac{(p+1)1^p\ln(1)+..+(p+1)(n)^p\ln(n)-n^{p+1}\ln(n)}{(p+1)n^{p+1}}$$
Then we apply the theorem of Stolz-Cesaro, as
\begin{align}\lim_{n\to\infty}&\frac{(p+1)(n+1)^p\ln(n)-(n+1)^{p+1}\ln(n+1)+n^{p+1}\ln(n)}{(p+1)((p+1)n^p+...+1)}\\&=\lim_{n\to\infty}\frac{(p+1)n^p\ln(n)-n^{p+1}\ln(n+1)-(p+1)n^p\ln(n)+n^{p+1}\ln(n)+o(n^p)}{(p+1)((p+1)n^p+o(n^p))}\\&=\lim_{n\to\infty}\frac{n\ln\left(1-\frac{1}{n+1}\right)+\frac{o(n^p)}{n^p}}{(p+1)((p+1)+\frac{o(n^p)}{n^p})}\\&=\frac{-1}{(p+1)^2}\end{align}
The various $o(n^p)$ are sum of the terms of the form $n^k\ln(n+1)$ or $n^k$, $k<n$ in the numerator and denominator respectively. It is clear why these terms are equal to $o(n^p)$
A: A couple thoughts here:
1) L'Hopital's rule might be helpful.  At the very least, it is worth exploring.
2) The advise to consider the log of the expression seems like a good idea, even
   if you do not use the integral to sum up the result.
If the limit of the natural log of the l.h.s expression is $-1/(p+1)^2$ then you have your result.
