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?
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 $f(x)=x^p,x\in[0,1]$)
