Bernouli's Formula for sum of kth powers of first n natural numbers is given by: $$f_k(n)=\frac{1}{k+1}\sum_{j=0}^k{k+1\choose j}B_j(n+1)^{k+1-j}$$ where $Bj$ is the $j^{th}$ Bernoulli Number and is in a sense recursively given by:$$B_j=-\frac{1}{j+1}\sum_{i=0}^{j-1}{j+1 \choose i}B_i$$.
I did find a generalized proof of this for Generalized case where powers can be complex numbers. I am looking for simpler proofs. Do you have any idea if this can be proved by induction.
Thank you.
PS. I am not sure of tags and appreciate if they are corrected.
Added By simpler I mean that do involve only integer powers.
The proof using the generating function: $$\frac{t}{e^t-1}$$ Can be found here. In essence, the proof follows from noting that: $$e^{kt}=\sum k^m \frac{t^m}{m!}$$ So that the sum $\ \sum k^m$ is related to the sum of the geometric series which is in turn related to the generating function.