Here's one way of proceeding: first, verify that both sides of the purported identity agree for $k=1$. Now, if you carry out integration by parts on the left hand side, you should obtain the recursion relation
$\beta_{k-1}=\beta_{k-2}+\frac{(\alpha w)^{k-1}e^{-\alpha w}}{(k-1)!}$
where
$\beta_{k-1}=\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\frac{\Gamma(k,\alpha w)}{\Gamma(k)}=Q(k,\alpha w)$
and $\Gamma(k,u)$ and $Q(k,u)$ are various notations for the (upper) incomplete gamma function.
The trick now is to replace $\beta_{k-1}$ in the recursion relation with the sum on the right hand side, and verify that the recursion still holds. This along with the verified initial condition $k=1$ proves your identity.
As a variation of Didier's strategy in the comments: after letting $\alpha w=0$, to verify the equation
$\int_0^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=e^{0}\left(1+\sum_{x=1}^{k-1} \frac{0^x}{x!}\right)$
one can demonstrate that the expression on the right does simplify to $1$; showing that
$(k-1)!=\int_0^\infty z^{k-1}e^{-z}\mathrm dz=\Gamma(k)$
can be done by verifying that both sides of the equation agree when $k=1$ and establishing the recursion $(k-1)!=(k-1)(k-2)!$ through integration by parts.
Differentiating both sides of the original identity with respect to $u=\alpha w$ yields the relation
$-\frac{u^{k-1}e^{-u}}{(k-1)!}=e^{-u}\left(\sum_{x=1}^{k-1} \frac{xu^{x-1}}{x!}-\sum_{x=0}^{k-1} \frac{u^x}{x!}\right)$
I'll leave the simplification to the reader.