Formula to obtain sum of $ 1^k+2^k+3^k+\cdots+(n-1)^k+n^k$

Question

How would one systematically obtain the formula for $$ 1^k+2^k+3^k+\cdots+(n-1)^k+n^k$$ for a fixed positive integer exponent k? I am aware of the formulas $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}\quad ,\quad\sum_{i=0}^n i = \frac{n(n+1)}{2}\quad,\quad\sum_{i=0}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$ i.e. when values of k are 1, 2 and 3. Using these can I use induction to proceed with the original sum?

Answer 1

A slightly different approach. One may use the Bernoulli polynomials satisfying $$B_{k+1}(x+1)-B_{k+1}(x)=(k+1)x^k$$ then, one gets $$\sum_{j=1}^n (k+1)j^k=\sum_{j=1}^n (B_{k+1}(j+1)-B_{k+1}(j))$$ but this is a telescoping sum giving

$$ \sum_{j=1}^n j^k=1^k+2^k+3^k+....+(n-1)^k+n^k=\frac{B_{k+1}(n+1)-B_{k+1}(1)}{k+1}. $$