Formula for $\sum_{i = 1}^n k^n$

Question

I know from my calculator the answer is $\sum_{i = 1}^n k^n$ = $\frac{k^{n+1}-k}{k - 1}$. I'd just like help understanding why.

Answer 1

Because $$(1-k)(1+k^2+k^3+\dots+k^n)=(1+k^2+k^3+\dots+k^n)-(k+k^2+k^3+\dots+k^{n+1})=1-k^{n+1}$$