<s>[INST] A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients 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 $\binom{n+1}{k}$ ?
 [/INST] 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,
$$
\sum_{x=1}^n x^5 = \binom{n+1}{2} +30\binom{n+1}{3} +150\binom{n+1}{4} +240\binom{n+1}{5} +120\binom{n+1}{6}
$$
</s>