sequences and series - Formula for $sum_{k=1}^{n}{k^p}$ where p is a positive integer

Any hints that can take me from here or am I completely lost.

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a-1}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{a1

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a-1}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{i=1}^{n}(n+1-i)^{a-1}(i)))$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{i=1}^{n}(n+1-i)^{a-1}(i)))$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(a!S(n,a))(\sum_{i=1}^{n}(i)^{a-1}(n+1-i)))$

Where S(n,a) is a stirling number of second kind.