I am trying to find the PDF of $Y$, the sum of I.I.D. exponential random variables $X_1, ... X_n$ with $\lambda = 1$ and $n$ some known constant.
So far, I have determined that moment-generating function/characteristic function $M_s(Y) = M_s(X_1)^n = \left( \dfrac{1}{1-s} \right)^n.$ But this does not match the characteristic function of any other simple RV so i am at a loss as to how to proceed.
If $X_1,X_2\sim\operatorname{Exp}(\lambda)$ are independent with densities $f$ and $g$ then the density of $X_1+X_2$ can computed as \begin{align} h_2(t) &= (f\star g)(t)\\ &= \int_0^t f(s)g(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \lambda^2 e^{-\lambda t} \int_0^t \ \mathsf dt\\ &= \lambda^2 t e^{-\lambda t}.\\ &= (\lambda t)\lambda e^{-\lambda t} \end{align} In general, $S_n:=\sum_{i=1}^n X_i$ has an Erlang distribution , with density $$h_n(t) = \frac{(\lambda t)^{n-1}\lambda e^{-\lambda t}}{(n-1)!}, $$ as $$h_1(t)=\lambda e^{-\lambda t} = \frac{(\lambda t)^{1-1}\lambda e^{-\lambda t}}{0!}\mathsf 1_{(0,\infty)}(t),$$ so by induction, \begin{align} h_{n+1}(t) &= (h_n\star h_1)(t)\\ &= \int_0^t \frac{(\lambda s)^{n-1}\lambda e^{-\lambda s}}{(n-1)!}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \frac{\lambda^{n+1} e^{-\lambda t}}{(n-1)!}\int_0^t s^{n-1}\ \mathsf ds\\ &= \left(\frac{\lambda^{n+1} e^{-\lambda t}}{(n-1)!}\right)\frac{t^n}n\mathsf 1_{(0,\infty)}(t)\\ &= \frac{(\lambda t)^{n}\lambda e^{-\lambda t}}{n!}\mathsf 1_{(0,\infty)}(t). \end{align} In the case where $\lambda=1$, this reduces to $$\frac{t^n e^{-t}}{n!}\mathsf 1_{(0,\infty)}(t). $$
The moment generating function of $S_n$ is \begin{align} M_{S_n}(s) &= \prod_{i=1}^n M_{X_i}(s)\\ &= M_{X_1}(s)^n. \end{align} As \begin{align} M_{X_1}(s) &= \mathbb E[e^{sX_1}]\\ &= \int_0^\infty e^{st}\lambda e^{-\lambda t}\ \mathsf dt\\ &=\int_0^\infty \lambda e^{-(\lambda-s)t}\ \mathsf ds\\ &=\frac{\lambda}{\lambda-s}, \end{align} we have $$ \mathbb E[e^{sS_n}] = \left(\frac{\lambda}{\lambda-s}\right)^n.$$
The sum you are looking at is a $Gamma(n;1)$ rv.
Here https://en.m.wikipedia.org/wiki/Gamma_distribution you can check your MGF and verify that it matches with the Gamma's one (or Erlang, as $n \in \mathbb{N}$
