While reviewing definite integrals, $\int_a^bf(x)dx$; I recalled that a definite integral could not only be solved by the difference of the anti-derivatives of intervals b and a, $F(b)-F(a)$, via the 2nd fundamental theorem of calculus, but also through the Riemann summation: $$\lim_{n\to\infty}f(x_i^*)\Delta x;\ \textrm{where}\ \Delta x=\frac{b-a}{n}\ \textrm{and}\ x_i^*=a+(\Delta x)i$$ Given this, I tried out an integral over an example polynomial. The difference of anti-derivatives would've given a sufficient answer, but I wanted to try it with Riemann sums. While attempting to solve this example, I came across sums of the form $\sum_{i=1}^ni^k$. Thankfully, I recalled the formulas for $k=\{1,2,3\}$ which were as follows: $$\sum_{i=1}^ni^1=\frac{n\cdot(n+1)}{2}$$ $$\sum_{i=1}^ni^2=\frac{n\cdot(n+1)\cdot(2n+1)}{6}$$ $$\sum_{i=1}^ni^3=\frac{n^2\cdot(n+1)^2}{4}$$ This allowed me to solve the problem, however I wondered about orders of k greater than 3. What is the general formula for $\sum_{i=1}^ni^k$ given any value of k and how would I prove it?
Who knows about the summation of this series: $$\sum\limits_{i=1}^{n}i^m $$  where $m$ is constant and $m\in \mathbb{N}$?  thanks
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