Question on gamma distribution and waiting times. Let $X_1, X_2,\dots$  be iid. random variables each with density $xe^{-x}$ for $x > 0$ and $0$
otherwise. Let $S_0 = 0$ and $S_n = X_1 +\cdots + X_n$, and $N(t) = \max\ \{n : S_n < t\}$.
I need to find the mass function of the random variable $N(t)$.
I have worked out that the $X_i$ are distributed with a gamma distribution with $n = 2$ and $\lambda = 1$. Subsequently $S_n$ is distributed with $\gamma(1,2n)$. I realise that somehow the gamma distribution as a waiting time means that $N(t)$ is a Poisson, but I cannot show it. Any help would be appreciated. 

Note that $$\{N(t)=n\}=\{S_n\geqslant t, S_{n-1}<t\}=\{X_n\geqslant t-S_{n-1},S_{n-1}<t \}. $$
Hence
\begin{align}
\mathbb P(N(t)=n) &= \mathbb P(X_n\geqslant t-S_{n-1},S_{n-1}<t)\\
&=\int_0^t\mathbb P(X_n\geqslant t-S_{n-1}\mid S_{n-1}=s)f_{S_{n-1}}(s)\ \mathsf ds\\
&= \int_0^t e^{-(t-s)}(1+(t-s))\frac{s^{2n-1}e^{-s}}{(2n-1)!}\ \mathsf ds\\
&= \frac{e^{-t}t^{2n}(2n+1+t)}{(2n+1)!}.
\end{align}
Alternatively, we have
$$\{N(t)=n\} = \{P(t)=2n\}\cup\{P(t)=2n+1\} $$
where $P(t)$ is a Poisson process with rate $1$, and so
\begin{align}
\mathbb P(N(t)=n) &= \mathbb P(P(t)=2n) + \mathbb P(P(t)=2n+1))\\
&= \frac{e^{-t}t^{2n}}{(2n)!} + \frac{e^{-t}t^{2n+1}}{(2n+1)!}\\
&= \frac{e^{-t}t^{2n}(2n+1+t)}{(2n+1)!}.
\end{align}
