The falling power $n^\underline{k}$ (read $n$ to the falling $k$) is defined as follows: $$n^\underline{k}=n(n-1)(n-2)\cdots(n-k+1)$$ These are important in discrete calculus because their finite differences and sums are analogous to those of normal powers in differential calculus: $$\Delta n^\underline{k} = k n^\underline{k-1}$$ It is easy to see that $n^1=n^\underline{1}$ and $n^2=n^\underline{2} + n^\underline{1}$. With these base cases, one can recursively express any normal power $n^k$ as a sum of falling powers of order $k$ and less by the following method: expand $n^\underline{k}$, yielding a $k$th order polynomial. Then, substitute falling-power-expansions for all terms of order less than $k$, expressing $n^\underline{k}$ as $n^k$ plus some lower-order falling powers. Rearrange to put $n^k$ on the LHS and all falling powers on the RHS. For example, one can find that $n^3 = n^\underline{3} + 3n^\underline{2} + n^\underline{1}$.
This is rather tedious to do by hand for large values of $k$. Is there a more direct method to expand $n^k$ into a sum of falling powers?
