This is a well-known result analogous to the formula $$ \frac{d^k}{dx^k} x^k = k!$$for differentiation of polynomials. I'll try to state it clearly:
Definition: Let the discrete derivative of a sequence $a_1, a_2, a_3, \ldots$ be the sequence $a'_n := a_{n+1} - a_n$.
Remark: The discrete derivative is a linear operator. That is, if $c_n = \alpha a_n + \beta b_n$ for some constants $\alpha$ and $\beta$, then $c'_n = \alpha a'_n + \beta b'_n$.
Notation: Iterated discrete derivatives can be notated by parenthesized superscripts; thus $c^{(2)}$ for $c''$, $c^{(3)}$ for $c'''$, and so on.
Theorem: For some $k \geq 1$ fixed, let $a_n = n^k$ be the sequence of $k$th powers. Then $a_n^{(k)} = k!$ for all $n$.
Remark: An immediate consequence of this statement, used in a proof by induction, is that $a_n^{(\ell)} = 0$ if $k < \ell$.
Proof. The result is obvious for $k = 1$. Now assume that the result holds for all integers less than $k$. Then $$a'_n = (n+1)^k - n^k = k n^{k-1} + \binom{k}{2} n^{k-2} + \binom{k}{3} n^{k-3} + \cdots + kn + 1.$$
We know that $a_n^{(k)}$ is the $k-1$th discrete derivative of the RHS. Because the discrete derivative is a linear operator, we can take the derivative of each term on the RHS by itself. We know by the induction hypothesis that the $k-1$th derivative of $n^{k-1}$ is $(k-1)!$ and that the $k-1$th derivative of every other term in the RHS is zero, and we're done.
Note that the $k$th discrete derivative is a constant $k!$ for any polynomial with leading term $n^k$, not just the $k$th powers, as terms of smaller order drop out.