Prove $\sum_{q=\alpha}^p \binom{q}{\alpha} \binom{p}{q}\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}.$

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

$$\frac{p!}{ r!}.$$

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.