TITLE: Gamma identity $\lim_{n\to \infty}n^{p+1}\int_{0}^{1}e^{-nx}\ln(1+x^{p})dx=\Gamma(p+1)$
QUESTION [6 upvotes]: I ran across what appears to be another Gamma identity. 

Show that $$\lim_{n\to \infty}n^{p+1}\int_{0}^{1}e^{-nx}\ln(1+x^{p}) \,\mathrm dx=\Gamma(p+1)=p!$$

I tried several different subs and even the series for $$\ln(1+x^{p})$$ and nothing materialized.
What would be a good start on this one?
It looks similar to $$\Gamma(p+1)=n^{p+1}\int_{0}^{\infty}x^{p}e^{-nx} \,\mathrm dx,$$ but I was unable to hammer it into that form. I would think there is a clever sub of some sort that may work it into this last mentioned form.

REPLY [2 votes]: Substituting $x\mapsto x/n$ yields
$$
\begin{align}
n^{p+1}\int_{0}^{1}e^{-nx}\log(1+x^{p})\,\mathrm{d}x
&=n^p\int_0^ne^{-x}\log(1+(x/n)^p)\,\mathrm{d}x\\
&=n^p\int_0^n\left((x/n)^p+O(x/n)^{2p}\right)e^{-x}\,\mathrm{d}x\\
&=\int_0^nx^pe^{-x}\,\mathrm{d}x+\int_0^\infty x^{2p}e^{-x}\,\mathrm{d}x\;O\left(n^{-p}\right)
\end{align}
$$
Thus,
$$
\begin{align}
\lim_{n\to\infty}n^{p+1}\int_{0}^{1}e^{-nx}\log(1+x^{p})\,\mathrm{d}x
&=\int_0^\infty x^pe^{-x}\,\mathrm{d}x\\
&=\Gamma(p+1)
\end{align}
$$