Calculate:   $$\lim_{n\to\infty} \frac{ (1^{1^p}2^{2^p}\cdot...\cdot n^{n^p})^{ 1/n^{p+1} }}{n^{1/(p+1)}}$$   I've done some steps as follows: $$a_n:=\frac{ (1^{1^p}2^{2^p}\cdot...\cdot n^{n^p})^{ 1/n^{p+1} }}{n^{1/(p+1)}} \iff \ln a_n=\frac{1}{n^{p+1}}\big(\sum_{k=1}^nk^p\ln k-\frac{n^{p+1}}{p+1}\ln n\big) \iff \\\ln a_n =\frac{1}{n}\sum_{k=1}^n\big[\big(\frac{k}{n}\big)^p\ln \frac{k}{n}\big]+\frac{1}{n}\sum_{k=1}^n\big(\frac{k}{n}\big)^p\ln n-\frac{\ln n}{p+1}.$$ Then, I was wondering if I could make some integrals out of it but still there are some odd terms.   I think my approach isn't so good...
Prove that for real number $p\geq  0$ we have   $$\lim_{n\to\infty}\frac{(1^{1^p} \cdot2^{2^p}\cdots n^{n^p})^\frac{1}{n^{p+1}}}{n^\frac{1}{(p+1)}} = \exp\left(\dfrac{-1}{(p+1)^2}\right).$$
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