Computing limit of a product Prove that:
$$\lim_{n \to \infty} n^{-1/(p+1)} (1^{1^p}2^{2^p} \cdots n^{n^p})^{1 /n^{p+1}} = e^{\frac{-1}{(p+1)^2}}$$
I really have no clue how to go about this. I try converting it into summation form by taking logarthims and then trying to approximate each term but to no avail. Can anyone please help me?
 A: Consider sequence
$$x_n=\log\frac{(1^{1^p}2^{2^p} \cdots n^{n^p})^{\frac{1}{n^{p+1}}}}{n^{\frac{1}{p+1}}}
= \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^p\log k-\frac{1}{p+1}\log n.$$
So $$x_n=\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^p\log \frac{k}{n}
+\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^p\log n
-\frac{1}{p+1}\log n.$$
For the first part we know that
$$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^p\log \frac{k}{n}
=\int_0^1 x^{p+1}\log x dx=\frac{-1}{(p+1)^2}.$$
And then you need prove that
$$\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^p\log n
-\frac{1}{p+1}\log n\to 0,$$
that is to say 
$$\lim_{n\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^p-\frac{1}{p+1}\right)\log n=0.$$

As we know that: If $f$ has continuous derivative on $[a,b]$, then 
  $$\lim_{n\to \infty}n\left(\frac{b-a}{n}\sum_{k=1}^{n}f\left(a+\frac{k(b-a)}{n}\right)
 -\int_{a}^{b}f(x)\mathrm{d}x\right)=\frac{f(b)-f(a)}{2}(b-a).$$
  This implies the above limit!(Take $f(x)=x^p,x\in[0,1]$)

