We are interested in a closed form of
$$\Omega_a(n)
= \sum_{k=0}^n (-1)^{n-k} {n\choose k} k^{n+a}$$
for natural $a.$ More precisely we seek $P_a(n)$ where
$$P_a(n) = \frac{\Omega_a(n)}{(n+a)!},$$ which is a
polynomial. Keeping it simple we find
$$(n+a)! [z^{n+a}]
\sum_{k=0}^n (-1)^{n-k} {n\choose k} \exp(kz)
\\ = (n+a)! [z^{n+a}] (\exp(z)-1)^n
= n! {n+a\brace n}.$$
With $a$ our constant we have combinatorially that we
are counting the number of partitions of $[n+a]$ into
an ordered sequence of $n$ sets. We can think of this
as first choosing $n$ values to go into our $n$ slots
for a factor of $n! {n+a\choose n}.$ Then we partition
the $a$ remaining values into a sequence of $q$ sets
where $1\le q\le a$ and attach them at ${n\choose q}$
points to the bottom layer of singletons. However we
cannot use $\exp(z)-1$ to select the sets that form the
partition of $a$ because we overcount by a factor of
$m$ on $z^m$. This is because if we select a set
containing $m$ elements the attachment point has $m$
possibilities. Hence we must divide by $m.$ Integrate
to get $\exp(z)-1-z$ and divide by $z$ for an EGF of
$(\exp(z)-1)/z-1.$ Putting it all together,
$$n! {n+a\choose n} \sum_{q=1}^a {n\choose q}
a! [z^a] ((\exp(z)-1)/z-1)^q
\\ = (n+a)! \sum_{q=1}^a {n\choose q}
[z^a] \frac{1}{z^q} (\exp(z)-1-z)^q.$$
We have our factorial in front which is canceled. Here
we already see that we have a polynomial in $n$ because
the factors on the binomial coefficients do not depend
on $n$ and the binomial coefficients yield polynomials
in $n$ upon expansion. Continuing,
$$\sum_{q=1}^a {n\choose q}
[z^{a+q}] \sum_{p=0}^{q} {q\choose p}
(-1)^{q-p} z^{q-p} (\exp(z)-1)^p
\\ = \sum_{q=1}^a {n\choose q}
\sum_{p=0}^{q} {q\choose p} p!
(-1)^{q-p} [z^{a+p}] \frac{(\exp(z)-1)^p}{p!}
\\ = \sum_{q=1}^a {n\choose q}
\sum_{p=0}^{q} {q\choose p} p!
(-1)^{q-p} \frac{1}{(a+p)!} {a+p\brace p}.$$
The conclusion is that
$$\bbox[5px,border:2px solid #00A000]{
P_a(n) = \frac{1}{a!} \sum_{q=1}^a {n\choose q}
\sum_{p=0}^{q} {q\choose p} (-1)^{q-p}
{a+p\choose p}^{-1}
{a+p\brace p}.}$$
We can compute an explicit form of the coefficients of
$P_a(n)$ by using
$$[n^r] {n\choose q}
= (-1)^{q+r} \frac{1}{q!} {q\brack r}$$
which then yields
$$\bbox[5px,border:2px solid #00A000]{
P_a(n) = \frac{1}{a!} \sum_{r=1}^a n^r (-1)^r
\sum_{q=1}^a \frac{1}{q!} {q\brack r}
\sum_{p=0}^{q} {q\choose p} (-1)^p
{a+p\choose p}^{-1}
{a+p\brace p}.}$$
The first sum in $r$ iterates over the powers of $n$ up
to $n^a$, forming a polynomial in $n$.
Remark. Once we have seen the closed form from
the combinatorial argument it becomes readily apparent
that we can also get it by simple algebraic
manipulation, as suggested in the comment, which goes
as follows. We have
$$(n+a)! [z^{n+a}] (\exp(z)-1)^n
= (n+a)! [z^{n+a}] \sum_{q=0}^n {n\choose q}
(\exp(z)-z-1)^q z^{n-q}
\\ = (n+a)! \sum_{q=0}^n {n\choose q}
[z^{a+q}] (\exp(z)-z-1)^q.$$
Note however that we may set the upper range to $a.$
This is because when $a\gt n$ the binomial coefficient
is zero in the added range and we are not adding
anything and may raise to $a$. On the other hand the
sum term starts at $z^{2q}$ so we get zero when $2q\gt
q+a$ or $q\gt a.$ Hence if $a\lt n$ we may lower to $a$
because we are not losing anything. We find
$$(n+a)! \sum_{q=1}^a {n\choose q}
[z^{a+q}] (\exp(z)-z-1)^q$$
and may then continue with the computation as
before. (The term for $q=0$ does not go past the
coefficient extractor.)
