Calculate $\lim_{n\to\infty} \frac{ (1^{1^p}2^{2^p}\cdot...\cdot n^{n^p})^{ 1/n^{p+1} }}{n^{1/(p+1)}}$ 
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...

Assuming $p > -1$, continuing your split up series.
1 .This is was WA gives us for the first summation. First sum, e.g. $p=3$ $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^p\ln \frac{k}{n} = -\frac{1}{(p+1)^2}$$. 
EDIT: Using the hint from @Riemann, this sum is equal to $\int_0^1x^p\ln(x)dx=-\frac{1}{(p+1)^2}$, for $p>-1$.


*Second $$\frac{\ln n}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^p - \frac{\ln n}{p+1}=\ln n\left[\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^p-\frac{1}{p+1}\right] = \ln n\left[\frac{1}{n^{p+1}}\sum_{k=1}^nk^p-\frac{1}{p+1}\right].$$
$\sum_{k=1}^nk^p$ can be written using Bernoulli numbers as $$\sum_{k=1}^nk^p=\frac{n^{p+1}}{p+1}+\frac{1}{2}n^p+\sum_{k=2}^p\frac{B_k}{k!}p^\underline{k-1}n^{p-k+1},$$
and thus we obtain $$\frac{1}{p+1}+\frac{1}{2}\frac{n^p}{n^{p+1}}+\frac{n^{p+1}}{n^{p+1}}\sum_{k=2}^p\frac{B_k}{k!}p^\underline{k-1}n^{-k}-\frac{1}{p+1}=$$
$$\frac{1}{2n}+\sum_{k=2}^p\frac{B_k}{k!}p^\underline{k-1}n^{-k}.$$
And therefore, since $\lim_{n\to\infty}\frac{\ln n}{n} = 0$, $$\lim_{n\to\infty}\frac{\ln n}{2n}+\sum_{k=2}^p\frac{B_k}{k!}p^\underline{k-1}\frac{\ln n}{n^{k}}=0$$
Combing the two series,
$$\lim_{n\to\infty}\ln a_n = -\frac{1}{(p+1)^2} + 0$$ and (to me somehow surprising)
$$\lim_{n\to\infty}a_n=e^{-\frac{1}{(p+1)^2}}$$
