Sums of Powers - A Festive Story

The general case:

\begin{eqnarray} 1^m +2^m + \dots + n^m \\ = \left.(1^me^t + 2^me^{2t} + \dots n^me^{nt})\right|_{t=0}\\ =\left. \frac{d^m}{dt^m}\left(1 +e^t + e^{2t} + \dots + e^{nt}\right) \right|_{t=0}\\ =\left. \frac{d^{m+1}}{dt^{m+1}}\frac{t(e^{(n+1)t} - 1)}{e^t-1} \right|_{t=0}\\ =\frac{1}{m+1}\left( \left. \frac{d^{m+1}}{dt^{m+1}}\frac{te^{(n+1)t}}{e^t-1}\right|_{t=0} -\left.\frac{d^{m+1}}{dt^{m+1}}\frac{t}{e^t-1}\right|_{t=0}\right) \\ =\frac{1}{m+1}\left( \left. \frac{d^{m+1}}{dt^{m+1}}\sum_{k=0}^{\infty}B_k(n+1)\frac{t^k}{k!}\right|_{t=0} -\left.\frac{d^{m+1}}{dt^{m+1}}\sum_{k=0}^{\infty}B_k\frac{t^k}{k!}\right|_{t=0}\right) \\ =\frac{B_{m+1}(n+1) - B_{m+1}}{m+1}\\ \end{eqnarray}

Where $B_n(x)$ is the Bernoulli polynomial and $B_n$ are the Bernoulli numbers

Conclude that $$1^m + 2^m + \dots + n^m = \frac{B_{m+1}(n+1) - B_{m+1}}{m+1}$$ Amongst other things, all you have to do now is find out what on earth is a Bernoulli polynomial!!