PROBLEM
Let $X_{1},\dots,X_{n}$ be independent and identically $\text{Exp}(1/\lambda)$-distributed random variables. Let in addition $S_{0}=0$ and $S_{n}=X_{1}+\dots+X_{n}$. Set
\begin{align} N = \max\{n\;\vert\; S_{n}\leq x\}. \end{align}
$N$ is a random time, equal to the number of that random sample, when $S_{n}$ for the last time stays under $x$. Then show that $N\in\text{Po}(\lambda x)$.
Here I define the exponential probability density, $\text{Exp}(\lambda)$, as
\begin{align} f(x) = \left\{\begin{matrix}\frac{1}{\lambda}e^{-\frac{x}{\lambda}} \quad 0\leq x,\\ 0\quad \text{elsewhere}. \end{matrix}\right. \end{align}
I have been told that characteristic functions would facilitate solving this problem, however these have not been introduced yet and I do not want to "cheat"! So, inspired by some other post on this forum I began computing the probabiltiy mass function for $N$, i.e. $p_{N}(n)$, for different values of $n$. We realize that $N$ is discrete and that $N\in \{0,\dots,n\}$. For $n=0$ I obtain
\begin{align} p_{N}(n=0) = \mathbb{P}(N=0) = \mathbb{P}(S_{1}>x) = 1-\mathbb{P}(X_{1}\leq x) = 1 - (1-e^{-\lambda x}) = e^{-\lambda x}. \end{align}
But if I try finding the probability mass function for $n=1$ I run into trouble,
\begin{align} p_{N}(n=1) = \mathbb{P}(N=1) = \mathbb{P}(\{S_{1}\leq x\},\{S_{2}>x\}) = \dots \end{align}
here I am unsure what to do, but I use the defintion of the conditional probabiltiy and obtain
\begin{align} \dots = \mathbb{P}(\{S_{2}>x\}\;\vert\; \{S_{1}\leq x\})\mathbb{P}(\{S_{1}\leq x\}), \end{align}
and from here I am not sure what to do. I can evaluate the second factor in the last equality, but the conditional probability I do not know how to obtain. The answer should be that $N\in\text{Po}(\lambda x)$, but how do I get to that (not using characteristic functions)?
The sum $S_n$ can be viewed as the time of the $n$th arrival in a Poisson process $M_t$ with arrival rate $\lambda$. Observe that
$$\left\{N = n\right\} = \left\{S_n \leq x,\, S_{n+1} > x\right\} = \left\{M_x=n\right\},$$
and $M_x\sim Poisson\left(\lambda x\right)$ by the properties of the Poisson process.