Prove $\sum_{q=\alpha}^p \binom{q}{\alpha} \binom{p}{q}\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}.$ How to prove 
$\displaystyle \sum_{q=\alpha}^p \binom{q}{\alpha} \binom{p}{q}\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}$
for $1 \leq \alpha \leq p$?
EDIT: This is a result that I derived after playing around with the (given) fact that
$\displaystyle\sum_{q=1}^p\sum_{j=1}^q q^{p-2}(1+h/q)^{j-1}\prod_{i=1,i\neq q}^p\frac{1}{q-i}=\sum_{q=1}^p[(1+h/q)^q-1]\frac{(-1)^{p-q}q^{p}}{q!(p-q)!}=\sum_{q=1}^p \frac{h^q}{q!}$
and grouping the coefficients of each $h^\alpha$ in both sides.
I have tried some values of $\alpha$ and $p$ and it works, so the formula seems to be true. I just don't know how to proceed to prove this result.
 A: Suppose we seek to verify that
$$(-1)^p \sum_{q= r}^p {p\choose q}
{q\choose r} (-1)^q q^{p- r} = \frac{p!}{ r!}.$$
We use the integral representation
$${q\choose  r} = {q\choose q- r}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon}
\frac{(1+z)^{q}}{z^{q- r+1}} \; dz$$
which is zero when $q\lt  r$  (pole vanishes) so we may extend $q$
back to zero. 
We also use the integral
$$q^{p- r}
= \frac{(p- r)!}{2\pi i} 
\int_{|w|=\gamma}
\frac{\exp(qw)}{w^{p- r+1}} \; dw.$$
We thus obtain for the sum
$$\frac{(-1)^p (p- r)!}{2\pi i} 
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}} 
\\ \times 
\frac{1}{2\pi i} 
\int_{|z|=\epsilon}
z^{ r-1}
\sum_{q=0}^p {p\choose q} 
(-1)^q \frac{(1+z)^q}{z^q} \exp(qw)
\; dz\; dw
\\ = \frac{(-1)^p (p- r)!}{2\pi i} 
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}} 
\\ \times 
\frac{1}{2\pi i} 
\int_{|z|=\epsilon}
z^{ r-1}
\left(1-\frac{1+z}{z}\exp(w)\right)^p
\; dz\; dw
\\ = \frac{(-1)^p (p- r)!}{2\pi i} 
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}}
\\ \times 
\frac{1}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{p- r+1}}
(-\exp(w)+z(1-\exp(w)))^p
\; dz\; dw
\\ = \frac{(p- r)!}{2\pi i} 
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}}
\\ \times 
\frac{1}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{p- r+1}}
(\exp(w)+z(\exp(w)-1))^p
\; dz\; dw.$$
We extract the residue on the inner integral to obtain
$$\frac{(p- r)!}{2\pi i} 
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}}
{p\choose p- r} \exp( r w)
(\exp(w)-1)^{p- r}
\; dw
\\ = \frac{p!}{ r!}
\frac{1}{2\pi i} 
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}}
\exp( r w)
(\exp(w)-1)^{p- r}
\; dw.$$
It remains to compute
$$[w^{p- r}]\exp( r w)
(\exp(w)-1)^{p- r}.$$
Observe  that $\exp(w)-1$  starts at  $w$  so $(\exp(w)-1)^{p- r}$
starts at $w^{p- r}$ and  hence only the constant coefficient from
$\exp( r  w)$  contributes, the  value  being  one, which  finally
yields
$$\frac{p!}{ r!}.$$
A: For my own convenience I’m going to replace your $\alpha,p$, and $q$ with $m,n$, and $k$, respectively. Fix $n$, and let
$$f(m)=\sum_k\binom{k}m\binom{n}k(-1)^{n+k}k^{n-m}=\sum_k(-1)^{n-k}\binom{n}kk^{n-m}\frac{k^{\underline{m}}}{m!}\;,$$
where $x^{\underline{m}}=x(x-1)\ldots(x-m+1)$ is a falling factorial. Now
$$\begin{align*}
\frac1{m+1}f(m)&=\sum_k(-1)^{n-k}\binom{n}kk^{n-m}\frac{k^{\underline{m}}}{(m+1)!}\\
&=\sum_k(-1)^{n-k}\binom{n}kk^{n-m-1}\frac{k\cdot k^{\underline{m}}}{(m+1)!}\\
&=\sum_k(-1)^{n-k}\binom{n}kk^{n-m-1}\frac{\big((k-m)+m\big)\cdot k^{\underline{m}}}{(m+1)!}\\
&=\sum_k(-1)^{n-k}\binom{n}kk^{n-m-1}\frac{k^{\underline{m+1}}+m\cdot k^{\underline{m}}}{(m+1)!}\\
&=f(m+1)+\sum_k(-1)^{n-k}\binom{n}kk^{n-m-1}\frac{m\cdot k^{\underline{m}}}{(m+1)!}\\
&=f(m+1)+\frac{m}{(m+1)!}\sum_k(-1)^{n-k}\binom{n}kk^{n-m-1}k^{\underline{m}}\;.
\end{align*}$$
Now $k^{n-m-1}k^{\underline{m}}$ is a polynomial in $k$ of degree $n-1$, say 
$$k^{n-m-1}k^{\underline{m}}=\sum_{i=0}^{n-1}c_ik^i\;,$$
so
$$\begin{align*}
\frac1{m+1}f(m)&=f(m+1)+\frac{m}{(m+1)!}\sum_k(-1)^{n-k}\binom{n}k\sum_{i=0}^{n-1}c_ik^i\\
&=f(m+1)+\frac{m}{(m+1)!}\sum_{i=0}^{n-1}c_k\sum_k(-1)^{n-k}\binom{n}kk^i\\
&=f(m+1)+\frac{m}{(m+1)!}\sum_{i=0}^{n-1}c_k{i\brace n}n!\\
&=f(m+1)\;,
\end{align*}$$
since the Stirling number of the second kind ${i\brace n}=0$ for $i<n$.
Now
$$f(0)=\sum_k\binom{n}k(-1)^{n-k}k^n={n\brace n}n!=n!\;,$$
so by an easy induction we have 
$$f(m)=\frac{n!}{m!}$$
for $0\le m\le n$.
I actually started from the observation that if in fact $f(m)=\frac{n!}{m!}$, then $f$ would have to satisfy the equation
$$\frac1{m+1}f(m)=f(m+1)$$
and worked from there.
A: Let $n:=p-\alpha$ and $r:=q-\alpha$.  Then, from the identity $\displaystyle\binom{q}{\alpha}\,\binom{p}{q}=\binom{p}{\alpha}\,\binom{p-\alpha}{q-\alpha}$, the equality $\displaystyle\sum_{q=\alpha}^p\,\binom{q}{\alpha}\,\binom{p}{q}\,\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}$ holds if and only if $$(-1)^n\,n!=\sum_{r=0}^n\,(-1)^r\,\binom{n}{r}\,(\alpha+r)^n\,.\tag{*}$$
For $j=0,1,2,\ldots,n$, the coefficient of $\alpha^j$ on the right-hand side is given by
$$t_j:=\sum_{r=0}^n\,(-1)^r\,\binom{n}{r}\,r^{n-j}\,,$$
where $0^0$ is set to be $1$.
Therefore, it suffices to show that $t_1=t_2=\ldots=t_n=0$ and $t_0=(-1)^n\,n!$.
This can be done, using induction on $j$ that $t_{n-j}=0$ if $j=0,1,\ldots,n-1$ and $t_n=(-1)^n\,n!$.  For $j=0$, the claim is trivial as $t_n=(1-1)^n$ which is $0$ if $n>0$, and which is $1$ if $n=0$.  Assume now that $n,j>0$ and $t_{n-j+1}=t_{n-j+2}=\ldots=t_{n}=0$.  Note that there are integers $d_1,d_2,\ldots,d_j$ such that $$r^{j}=j!\,\binom{r}{j}+d_{1}\,\binom{r}{j-1}+\ldots+d_j\,\binom{r}{0}$$
for all $r\in\mathbb{Z}$. Thus,
$$t_{n-j}=j!\,\sum_{r=0}^n\,(-1)^r\,\binom{n}{r}\,\binom{r}{j}+\sum_{i=1}^j\,d_i\,t_{n-j+i}\,.$$
By the induction hypothesis, $\displaystyle t_{n-j}=j!\,\sum_{r=0}^n\,(-1)^r\,\binom{n}{r}\,\binom{r}{j}$.  Consequently, 
$$t_{n-j}=j!\,\sum_{r=0}^n\,(-1)^r\,\binom{n}{j}\,\binom{n-j}{r-j}=(-1)^j\,j!\,\binom{n}{j}\,\sum_{r=j}^n\,(-1)^{r-j}\,\binom{n-j}{r-j}\,.$$
Ergo,
$$t_{n-j}=(-1)^j\,j!\,\binom{n}{j}\,\sum_{\mu=0}^{n-j}\,(-1)^\mu\,\binom{n-j}{\mu}=(-1)^j\,j!\,\binom{n}{j}\,(1-1)^{n-j}$$
which is $0$ if $j<n$, and is $(-1)^n\,n!$ if $j=n$.  The induction is now complete.
We have shown that (*) holds for any nonnegative integer $\alpha$, whence the original identity is also true.  Note that the identity (*) holds in $\mathbb{Z}[\alpha]$, where $\alpha$ is treated as a variable.
P.S.:  We don't need the fact that the $d_i$'s are integers.  It is sufficient to show that the $d_i$'s are rational numbers.  Simply observe that $\mathbb{Q}[x]$ is spanned by polynomials of the form $\binom{x}{i}$ for $i=0,1,2,\ldots$.  Then, $x^j\in\mathbb{Q}[x]$ is linear combination over $\mathbb{Q}$ of $\binom{x}{i}$ for $i=0,1,\ldots,j$.  However, it can be easily shown that $x^j$ is indeed a linear combination over $\mathbb{Z}$ of $\binom{x}{i}$ for $i=0,1,\ldots,j$.

Alternatively, observe that $\displaystyle s_j:=\sum_{r=0}^n\,(-1)^{n-r}\,\binom{n}{r}\,r^j=(-1)^n\,t_{n-j}$ is the number of surjections from $\{1,2,\ldots,j\}$ to $\{1,2,\ldots,n\}$, using the Principle of Inclusion and Exclusion.  Clearly, if $j=0,1,\ldots,n-1$, $(-1)^n\,t_{n-j}=s_j=0$ because such surjections do not exist, whereas $s_n=(-1)^n\,t_0$ is the number of permutations on $\{1,2,\ldots,n\}$, which is precisely $n!$.  
