Suppose that $N_t$ is a Poisson process, so that $$f(t, n) := P\{N_t = n\} = \frac{e^{-\lambda t} (\lambda t)^n}{n!}.$$
How do I compute the density of the following event $$S_n = \inf \{t \geq 0 : N_t = n\}?$$
One thing I know is $\{S_n = s\}$ is the same as $\{N_s = n \land N_t < n, \forall t < s\}$. Can someone help me?
For every $\delta\in(0,s)$ we have:$$S_n=s\implies N_s-N_{s-\delta}>0$$
so that: $$P(S_n=s)\leq P(N_s-N_{s-\delta}>0)=1-P(N_s-N_{s-\delta}=0)=1-e^{-\lambda\delta}$$
Let $\delta$ approach to $0$ and draw conclusions.
addendum (because of update of question)
$$P(S_n>t)=P(N_t<n)=\sum_{k=0}^{n-1}e^{-\lambda t}\frac{(\lambda t)^k}{k!}$$
This enables you to find the CDF and (by differentiating) also the PDF of $S_n$
extra info (nice to know about Poisson processes).
If $X_1,X_2,\dots$ are iid with exponential distribution equipped with parameter $\lambda$ then they "induce" a Poisson process $N_t$ with rate $\lambda$.
This by stating that $S_n:=\sum_{k=1}^nX_k$ and defining:
$$N_t:=|\{k\in\mathbb N\mid S_k\leq t\}|$$
So the density of $S_n$ could also be found on base of $S_n:=\sum_{k=1}^nX_k$.