A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients

Question

It is known that $$ \sum_{k=1}^{n}k^1=\binom{n+1}{2} $$ and $$ \sum_{k=1}^{n}k^2=\binom{n+1}{2}+2\binom{n+1}{3} $$ Is there a formula for $$ \sum_{k=1}^{n}k^m, $$ where $m$ is a positive integers, using binomial coefficients of the form $\binom{n+1}{k}$ ?

Answer 1

Note $$ x^m = \sum_{k=0}^m S_{m,k}\binom{x}{k} \tag1$$ where $$ S_{m,k} = S_2(m,k) k! $$ and $S_2(m,k)$ is the Stirling number of the second kind.

For $m \ge 1$ we have $S_2(m,0) = 0$, so $$ x^m = \sum_{k=1}^m S_{m,k}\binom{x}{k} \tag2$$ For example $$ x^5 = \binom{x}{1} +30\binom{x}{2} +150\binom{x}{3} +240\binom{x}{4} +120\binom{x}{5} $$

From $(2)$ we get $$ \sum_{x=1}^{n-1} x^m = \sum_{x=1}^n \sum_{k=1}^m S_{m,k}\binom{x}{k} = \sum_{k=1}^m S_{m,k} \sum_{x=1}^n \binom{x}{k} \tag3$$ Now $$ \sum_{x=1}^n \binom{x}{k} = \binom{n+1}{k+1}\qquad\text{if } k \ge 1 $$ so from $(3)$ we have our answer: if $m \ge 1$ then $$ \sum_{x=1}^{n} x^m = \sum_{k=1}^m S_{m,k} \binom{n+1}{k+1} \tag4$$

Recall: $S_{m,k} = S_2(m,k) k!$ where $S_2(m,k)$ is the Stirling number of the second kind.

For example, $$ \sum_{x=1}^n x^5 = \binom{n+1}{2} +30\binom{n+1}{3} +150\binom{n+1}{4} +240\binom{n+1}{5} +120\binom{n+1}{6} $$