Compute limit of double exponent expression. Compute $$\large\lim_{n\to \infty}\left(\left(\prod_{i=1}^{n+1}{i^{i^p}}\right)^{1/(n+1)^{p+1}} - \left(\prod_{i=1}^{n}{i^{i^p}}\right)^{1/n^{p+1}}\right),$$
where $p$ is some nonnegative real number.
I found this problem while working through the Problems in Real Analysis: Advanced Calculus on the Real Axis by Teodora-Liliana Radulescu, Titu Andreescu, and Vicenţiu D. Rădulescu, but I don't know how to do it. Can someone help me?

The calculations got a bit messy so there could be errors, let me know if you find some.
Let the sequence $a_n$ be defined by $a_n = \prod\limits_{i=1}^n i^{i^p}$.
Then by Abel's summation formula we have $$\log \left(a_n^{1/n^{p+1}}\right) = \frac{1}{n^{p+1}}\sum\limits_{i=1}^ni^p\log i = \frac{1}{n^{p+1}}\left(n^pS(n) - p\int\limits_1^nx^{p-1}S(x) \text{d}x\right)$$
where $S(x) = \sum\limits_{1 \leq i \leq x}\log i = x \log x - x + \mathcal{O}\left(\log x\right)$.
The expression becomes messy: $$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) - \frac{p}{n^{p+1}}\int\limits_1^nx^p\left(\log x - 1 + \mathcal{O}\left(\frac{\log x}{x} \right)\right)\text{d}x = $$
$$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) - \frac{p}{n^{p+1}}\int\limits_1^nx^p\left(\log x - 1\right)\text{d}x + \mathcal{O}\left(\frac{p}{n^{p+1}}\int\limits_1^nx^{p-1}\log x \text{d}x\right) = $$
$$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) - \frac{p}{n^{p+1}}\left(\frac{n^{p+1}\left((p+1)\log n - 1\right)+1}{(p+1)^2} - \frac{n^{p+1}-1}{p+1}\right) + \mathcal{O}\left(\frac{\log n}{n}\right) = $$
$$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) -\frac{p}{p+1}\log n + \frac{p}{(p+1)^2} + \frac{p}{p+1} + \mathcal{O}\left(\frac{\log n}{n}\right) = $$
$$\frac{\log n}{p+1}-\frac{1}{(p+1)^2}+\mathcal{O}\left(\frac{\log n}{n}\right)$$
So we get $$a_n^{1/n^{p+1}} = \left(1+\mathcal{O}\left(\frac{\log n}{n}\right)\right)e^{-1/(p+1)^2}n^{1/(p+1)}$$
Then we have $$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = e^{-1/(p+1)^2}\left((n+1)^{1/(p+1)}-n^{1/(p+1)}\right) + \mathcal{O}\left(n^{-p/(p+1)}\log n\right)$$
So we finally get
$$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = \mathcal{O}\left(n^{-p/(p+1)}\log n\right)$$
which converges to $0$.

I just realized that the last statement holds only for $p \gt 0$. If $p = 0$, the bound is not strong enough, and we need a better approximation.
When $p = 0$, the formula for $\log\left(a_n^{1/n^{p+1}}\right)$ becomes:
$$\log\left(a_n^{1/n^{p+1}}\right) = \frac{1}{n}\sum\limits_{i=1}^n\log i = \frac{1}{n}S(n)$$
and we use the better estimate $$S(n) = \sum\limits_{i=1}^n\log i = n\log n - n + \frac{1}{2}\log n + C + \mathcal{O}\left(\frac{1}{n}\right)$$ coming from Euler-Maclaurin summation (the constant value is $C = \log\sqrt{2\pi}$ but this is not necessary here).
Then we get $$\log\left(a_n^{1/n^{p+1}}\right) = \log n - 1 + \frac{1}{2}\frac{\log n}{n} + \frac{C}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)$$
and so $$a_n^{1/n^{p+1}} = \frac{n}{e}\left(1 + \frac{\log n}{2n} + \mathcal{O}\left(\frac{\log^2 n}{n^2}\right)\right)\left(1 + \frac{C}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)\right)\left(1 + \mathcal{O}\left(\frac{1}{n^2}\right)\right)$$
$$a_n^{1/n^{p+1}} = \frac{1}{e}\left(n + \frac{1}{2}\log n + C\right) + \mathcal{O}\left(\frac{\log^2 n}{n}\right)$$
Then we have $$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = \frac{1}{e}\left(1 + \frac{1}{2}\log\left(1+\frac{1}{n}\right)\right) + \mathcal{O}\left(\frac{\log^2 n}{n}\right)$$
$$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = \frac{1}{e} + \mathcal{O}\left(\frac{\log^2 n}{n}\right)$$
so the limit is $\frac{1}{e}$ in the case where $p = 0$.
