How is the Erlang pdf derived? If each arrival is exponentially distributed, then the $k$th arrival time is Erlang distributed. The Erlang PDF is:
$$
f_{Y_k}(y) = \lambda e^{-\lambda y} \frac{(\lambda y)^{k-1}}{(k-1)!}
$$
How is this derived?
 A: You can compute this by means of a convolution of $X \sim Exp(\lambda)$ and $Y_k \sim Erlang(k,\lambda)$ resulting in the distribution of $Y_{k+1} = X + Y_k$. (see also Why is the sum of two random variables a convolution?)
$$\begin{array}{rcccl}
f_{Y_{k+1}}(s) &=&  \int_0^s  \underbrace{ \vphantom{ \frac{(\lambda t)^{k-1}}{(k-1)!}}  \lambda e^{-\lambda (s-t)}}_{\text{exponential density}} \cdot \underbrace{ \lambda^k e^{-\lambda t} \frac{t^{k-1}}{(k-1)!}}_{\text{Erlang density with $k$}} dt && \\
& =& \lambda^{k+1} e^{-\lambda s} \int_{0}^s      \frac{t^{k-1}}{(k-1)!}  dt &=& \underbrace{ \vphantom{ \frac{(\lambda t)^{k-1}}{(k-1)!}} \lambda^{k+1} e^{-\lambda s} \frac{s^{k}}{k!}}_{\text{Erlang density with $k+1$}} 
\end{array}$$ 
That a sum of $k$ identical and independent exponential distributed variables is Erlang distributed will follow by induction.

Alternatively you can use Fourier transform or the characteristic function to quickly observe that characteristic function of the Erlang distribution $\varphi(t)_{\text{Erlang}} = \left(1 - \frac{it}{\lambda} \right)^{-k}$ is a product of the characteristic function of the exponential distribution $\varphi(t)_{exponential} = \left(1 - \frac{it}{\lambda} \right)^{-1}$. 
