TITLE: Showing $ \sum_{k=p} ^n {n-p \choose k-p} \frac{p^k}{k^p} \ge \frac{(p+1)^n}{(n+1)^p}$ for $0\leq p\leq n$
QUESTION [8 upvotes]: Let $n$ be a natural number, and $ 0 \le p\le n$. How can the following be shown?
$$ \sum_{k=p} ^n {n-p \choose k-p} \frac{p^k}{k^p} \ge \frac{(p+1)^n}{(n+1)^p}$$

Both sides are obviously equal when $p=0$ or $p=n$.
I have observed by numerics that the left-hand side seems to exceed the right-hand side but by little, not more than $17$%, when $p$ is close to $\frac{n}{2}$, as soon as $n$ is large enough.
Context:  a proof of the above inequality would enable an induction proof for this other inequality.
Update: From @Andreas partial answer, we see that a proof would follow if one could show that  $ p \mapsto   f(y(p),p)$ is a concave function of the variable $p$, in the range $[1,n]$, with
$$y(p):= e^{\frac{p-1}{n}}$$ and
$$ f(y,p):= \Big(y^{-\frac{y}{y+p}} \frac{p+y}{p+1} \Big)^n \Big(y^{\frac{y}{y+p}} \frac{n+1}{n+y}\Big)^p$$
since $f(y(1),1)=f(y(n),n)=1$, we would then have $f(y(p),p) \ge 1$ for $1\le p \le n$.
But this far from easy because the second derivative of $f(y(p),p)$ is awful.

REPLY [3 votes]: Using the integral representation obtained by Andreas we want to show $$\frac{1}{\Gamma(p)}\frac{p^n}{n^p} \int_{0}^{\infty} x^{p-1} e^{-x}  \left( 1 + \frac{e^{\frac{x}{n}}}{p} \right)^{n-p}  {\rm d}x \geq \frac{(p+1)^n}{(n+1)^p}$$
that is, after substituting $x=\frac{n}{n+1}u$ and using $$\left(1+\frac{e^{\frac{u}{n+1}}}{p}\right)^{n-p} \geq \left(1+\frac{1}{p}\right)^{n-p} e^{\frac{(n-p)u}{(n+1)(p+1)}}$$
for $n\geq p> 0$
$$\frac{1}{\Gamma(p)}\frac{p^n}{(n+1)^p} \int_{0}^{\infty} u^{p-1} e^{-\frac{n}{n+1}u}  \left( 1 + \frac{e^{\frac{u}{n+1}}}{p} \right)^{n-p} \,  {\rm d}u \\
\geq \frac{1}{\Gamma(p)}\frac{p^p(p+1)^{n-p}}{(n+1)^p} \int_{0}^{\infty} u^{p-1} e^{-\frac{n}{n+1}u}  e^{\frac{(n-p)u}{(n+1)(p+1)}} \, {\rm d}u \\
=\frac{1}{\Gamma(p)}\frac{p^p(p+1)^{n-p}}{(n+1)^p} \frac{(p+1)^p}{p^p} \, \Gamma(p) = \frac{(p+1)^n}{(n+1)^p} \, .$$