# How to derive Erlang distribution from the Exponential distribution?

I read that for a random variable $X$ to have Erlang distribution, it will be the sum of identical random variables with exponential distribution, but i cant derive the formula.

The density of an Erland distribution is $f(x;k,\lambda) = {\lambda ^k x^{k-1} e^ {- \lambda x} \over (k-1)!}$.