Showing $ \sum_{k=p} ^n {n-p \choose k-p} \frac{p^k}{k^p} \ge \frac{(p+1)^n}{(n+1)^p}$ for $0\leq p\leq n$ 
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.
 A: Rewrite $I$, the LHS in the task, as:
$$
I = \sum_{k=p} ^n {n-p \choose k-p} \frac{p^k}{k^p} =\sum_{k=p} ^n {n-p \choose n - k} \frac{p^k}{k^p} = \\
= \sum_{m=0}^{n-p}\frac{p^{n-m}}{(n-m)^p} {n-p \choose m} = \frac{p^n}{n^p}\sum_{m=0}^{n-p}\frac{p^{-m}}{(1-\frac{m}{n})^p} {n-p \choose m} $$
Now notice the following integral representation, derived from tabular integration by parts, which can be found in (only the German) Wikipedia here:
$$
\frac{1}{a^p} = \frac{1}{(p-1)!}\int_{x=0}^{\infty} x^{p-1} e^{-a x} dx \tag{1}
$$
Using that representation, we can identify $a = 1 - \frac{m}{n}$ and solve the following sum:
$$
S(x) = \frac{p^n}{n^p}\sum_{m=0}^{n-p}{p^{-m}}e^{\frac{m}{n}x} {n-p \choose m} $$
and then retrieve the result by integrating:
$$
I = \frac{1}{(p-1)!}\int_{x=0}^{\infty} x^{p-1} e^{-x} S(x) dx \tag{2}
$$
We obtain $S(x)$ by the Binomial theorem:
$$
S(x) = \frac{p^n}{n^p} \Big[ 1 + \frac{e^{\frac{x}{n}}}{p} \Big]^{n-p}\\
I = 
\frac{1}{(p-1)!}\frac{p^n}{n^p} \int_{x=0}^{\infty} x^{p-1} e^{-x}  \Big[ 1 + \frac{e^{\frac{x}{n}}}{p} \Big]^{n-p}  dx  $$
where $I$ were a full result after integration. Unfortunately, I do not see how the integration can be performed, so instead I considered a lower bound. The arguments are as follows:
The integral kernel in (2) is $x^{p-1} e^{-x} $ which has its maximum at $x = p-1$, so the "most important" part of the integrand is about that $x$ and any approximation of $S(x)$ should be exact there.
We therefore seek to approximate, in $S(x)$, the base $f(x) = 1 + \frac{e^{\frac{x}{n}}}{p}$ by a lower bound $g(x) = b e^{c x}$ with the properties $f(x=x_0) = g(x=x_0)$ and $f'(x=x_0) = g'(x=x_0)$. We can then choose $x_0 = p-1$ to meet the condition above, see the discussion below.
Straightforward calculation gives
$$
c = \frac{\frac{1}{pn} e^{\frac{x_0}{n}}}{1 + \frac{e^{\frac{x_0}{n}}}{p} } \quad ; \quad b = \frac{1 + \frac{e^{\frac{x_0}{n}}}{p} }{e^{c x_0}}
$$
Note that, comparing the exponents in $f(x)$ and $g(x)$,  indeed $c < \frac{1}{n}$ so $g(x)$ is a lower bound to $f(x)$ for all $x \ge 0$.
Let us first remark that an idea not followed here is to set $x_0 = 0$. This gives $c = \frac{1}{n (p+1)}$ and $b = 1 +  \frac{1}{p}$, which leads to the approach followed in Diger's answer.
Setting, as discussed above,   $x_0 = p-1$, leads to
$$
c = \frac{\frac{1}{pn} e^{\frac{p-1}{n}}}{1 + \frac{e^{\frac{p-1}{n}}}{p} } \quad ; \quad b = \frac{1 + \frac{e^{\frac{p-1}{n}}}{p} }{e^{c(p-1)}}
$$
Continuing with these values, we can now write a lower bound for the integration (2):
$$
I > 
\frac{1}{(p-1)!}\frac{p^n}{n^p} \int_{x=0}^{\infty} x^{p-1} e^{-x} \Big[ be^{cx} \Big]^{n-p}   dx = 
\frac{b^{n-p}}{(p-1)!}\frac{p^n}{n^p} \int_{x=0}^{\infty} x^{p-1} e^{-x(1- c(n-p))}  dx
$$
Again, it can be verified directly that $1- c(n-p) > 0$. We can now use the formula (1) again, this time identifying $a = 1- c(n-p)$. This results in
$$
I > 
\frac{b^{n-p}}{(1-c(n-p))^p}\frac{p^n}{n^p}  = \frac{(p+1)^{n}}{(n+1)^{p}}\frac{\big(\frac{p}{p+1} \big)^n}{\big(\frac{n}{n+1}\big)^p} 
\frac{b^{n-p}}{(1-c(n-p))^p}
$$
So in order to establish  the given inequality $
I >  \frac{(p+1)^{n}}{(n+1)^{p}}
$ we need that
$$
I_2 = \frac{\big(\frac{p}{p+1} \big)^n}{\big(\frac{n}{n+1}\big)^p} 
\frac{b^{n-p}}{(1-c(n-p))^p} \ge  1
$$
Inserting $b$ and $c$ we find that $I_2 = 1$ holds with equality for $\{p=1 \; ; \; \forall n\}$ and for $p=n$. Let us write $p = 1 + q (n-1)$, then if we show that $I_2$ holds for all $n$ and for $0 \leq q \leq 1$, we are done.
I haven't done this yet, but it is just a remaining algebraic task. Numerical simulations show that it is true, illustrated in the following contour plots of $I_2$, for small $n$ and larger $n$:


The maximum deviation is about 16 %. Since OP has stated in the task that the original maximum deviation is about 17 %, this shows that the above  lower bound to $S(x)$ almost doesn't change the integral $I$,  verifying the arguments for the chosen approximation.
A: 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 $n\geq p> 0$
$$\frac{1}{\Gamma(p)}\frac{p^n}{(n+1)^p} \int_{0}^{\infty} u^{p-1} e^{-\frac{n}{n+1}u}  \left( 1 + \frac{e^{\frac{u}{n+1}}}{p} \right)^{n-p} \,  {\rm d}u \\
\geq \frac{1}{\Gamma(p)}\frac{p^p(p+1)^{n-p}}{(n+1)^p} \int_{0}^{\infty} u^{p-1} e^{-\frac{n}{n+1}u}  e^{\frac{(n-p)u}{(n+1)(p+1)}} \, {\rm d}u \\
=\frac{1}{\Gamma(p)}\frac{p^p(p+1)^{n-p}}{(n+1)^p} \frac{(p+1)^p}{p^p} \, \Gamma(p) = \frac{(p+1)^n}{(n+1)^p} \, .$$
