binomial expression of a powered term One answer to a previous question of mine asserted that $$k^2=\binom k2+\binom {k+1}2.$$ 
I checked that the formula is true. However, it intrigued me. Is there a similar expression for $k^3$? How would I find a binomial for $k^n$? This is not a duplicate question to the best of my knowledge.

I'm not sure if there's a good way to do this in the general case, but just playing around with $k^3$, you know you need something with $\binom{k}{3}$ to get the cubic term. So:
$$\begin{split}
\binom{k}{3} &= \frac{k(k-1)(k-2)}6 = \frac{k^3-3k^2+2k}6 \\
\binom{k+1}{3} &= \frac{k(k+1)(k-1)}6 = \frac{k^3-k}6 \\
\binom{k+2}{3} &= \frac{k(k+2)(k+1)}6 = \frac{k^3+3k^2+2k}6
\end{split}$$
Adding the first and third lines lets you get rid of the $k^2$ term:
$$\binom{k}{3} + \binom{k+2}{3} = \frac{2k^3 + 4k}6$$
And then adding in 4 times the second line lets you get rid of the $k$ term:
$$\binom{k}{3} + 4\binom{k+1}{3} + \binom{k+2}{3} = k^3$$
Through similar fiddling, I can come up with:
$$k^4 = \binom{k}{4} + 11\binom{k+1}{4} + 11\binom{k+2}{4} + \binom{k+3}{4}$$
Though I do not know what the pattern is here. 
