$n(n-1)...(n-(k-1))$ is the number of injective functions from a set of size $k$ to a set of size $n$. We can count these using inclusion-exclusion: first include all such functions, of which there are $n^k$. Then, for each transposition $(ij)$ in $S_k$, exclude all the functions such that $f(i) = f(j)$, of which there are $n^{k-1}$. And so forth.

This argument is a little boring to make rigorous because there's a much easier way to prove an equivalent identity, which is

$$\frac{1}{k!} \sum_{\pi \in S_n} n^{\text{cyc}(\pi)} = {n+k-1 \choose k}.$$

This identity is equivalent because the sign of a permutation is determined by the parity of its number of cycles, and it corresponds to replacing "antisymmetric" by "symmetric" everywhere in your question. But this identity has an obvious proof by Burnside's lemma: the LHS and RHS both count the number of orbits of functions $[k] \to [n]$ under permutation of the domain. (This is a special case of a result I call the baby Polya theorem.)

Both identities are in turn a special case of the exponential formula, which one can state as a generating function identity for the cycle index polynomials of the symmetric groups. I explain some of this here.

$n(n-1)...(n-(k-1))$ is the number of injective functions from a set of size $k$ to a set of size $n$. We can count these using inclusion-exclusion: first include all such functions, of which there are $n^k$. Then, for each transposition $(ij)$ in $S_k$, exclude all the functions such that $f(i) = f(j)$, of which there are $n^{k-1}$. And so forth. This alternating sum cancels out all functions which are invariant under a permutation of the domain, so the only ones left are the injective ones.

There's a much easier way to prove an equivalent identity, which is

$$\frac{1}{k!} \sum_{\pi \in S_n} n^{\text{cyc}(\pi)} = {n+k-1 \choose k}.$$

This identity is equivalent because the sign of a permutation is determined by the parity of its number of cycles, and it corresponds to replacing "antisymmetric" by "symmetric" everywhere in your question. But this identity has an obvious proof by Burnside's lemma: the LHS and RHS both count the number of orbits of functions $[k] \to [n]$ under permutation of the domain. (This is a special case of a result I call the baby Polya theorem.)

Both identities are in turn a special case of the exponential formula, which one can state as a generating function identity for the cycle index polynomials of the symmetric groups. I explain some of this here. The relevant specializations are

$$\frac{1}{(1 - t)^n} = \exp \left( nt + \frac{nt^2}{2} + ... \right)$$

and

$$(1 + t)^n = \exp \left( nt - \frac{nt^2}{2} \pm ... \right).$$