Compute the distribution of $S_{N_t}$ 
In a Poisson process, we define $S_n$ is the arrival epoch of the $n$th occurence, $N_t$ is the number of occurence in $(0,t]$. Compute the distribution of $S_{N_t}$.

I made many trials to this problem. We need to compute the conditional probability of $S_k\le u$ given $N_t=k$. 
We have $P(N_t=k)=\frac{(\lambda t)^k}{k!}e^{-\lambda  t}$, thus we only need to compute $P(S_k\le u, N_t=k)$. 
The event $\left\{S_k\le u\right\}$ is the same as $\left\{N_u\ge k\right\}$, thus $P(S_k\le u, N_t=k)=P(N_u- N_t\ge 0)=1-P(N_t-N_u>0)$. Note that $N_t-N_u>0$ implies $t>u$, thus $P(N_t-N_u>0)=P(N_{t-u}>0)=1-P(N_{t-u}=0)$, and we can compute this probability easily.
I'm not sure whether my proof is correct. I will appreciate any help in case my proof is incorrect. Thanks in advance.
 A: Fix $t\in [0,\infty)$, and note that for $x \geq 0$, we have that
\begin{align*}
P(S_{N_t}\leq x) 
&= \sum_{k=0}^\infty P(S_{N_t}\leq x,N_t=k) \\
&=\sum_{k=0}^\infty P(S_{k}\leq x,N_t=k) \\
&=\sum_{k=0}^\infty P(N_x\geq k,N_t=k) \\
&=1_{\{x\geq t\}}(x)\sum_{k=0}^\infty P(N_{x}-N_{t}\geq 0,N_t=k)+ 1_{\{x< t\}}(x)\sum_{k=0}^\infty P(N_{t}-N_{x}\leq 0,N_t=k)\\
&=1_{\{x\geq t\}}(x)\sum_{k=0}^\infty P(N_{x}-N_{t}\geq 0,N_t=k)+ 1_{\{x< t\}}(x)\sum_{k=0}^\infty P(N_{t}-N_{x} = 0, N_t=k)\\
&=1_{\{x\geq t\}}(x)\sum_{k=0}^\infty P(N_{x}-N_{t}\geq 0,N_t=k)+ 1_{\{x< t\}}(x)\sum_{k=0}^\infty P(N_t-N_{x} = 0, N_x=k)\\
&= (*)
\end{align*}
Note that since $(N_t)$ has independent increments we have that for $x\leq y $ that $N_y -N_x$ is independent of $N_x=N_x-N_0$. Therefore 
\begin{align*}
(*) &=1_{\{x\geq t\}}(x)\sum_{k=0}^\infty P(N_{x}-N_{t}\geq 0)P(N_t=k)+ 1_{\{x< t\}}(x)\sum_{k=0}^\infty P(N_{t}-N_{x} = 0) P(N_x=k)\\
&=1_{\{x\geq t\}}(x)\sum_{k=0}^\infty P(N_{x-t}\geq 0)P(N_t=k)+ 1_{\{x< t\}}(x)\sum_{k=0}^\infty P(N_{t-x}= 0) P(N_x=k)\\
&=1_{\{x\geq t\}}(x)\sum_{k=0}^\infty P(N_t=k)+ 1_{\{x< t\}}(x)\sum_{k=0}^\infty \frac{(\lambda( t-x))^0}{0!}e^{-\lambda (t-x)} \frac{(\lambda x)^k}{k!}e^{-\lambda x}\\
&=1_{\{x\geq t\}}(x)+ 1_{\{x< t\}}(x) e^{-\lambda (t-x)} \sum_{k=0}^\infty  \frac{(\lambda x)^k}{k!}e^{-\lambda x}\\
&=1_{\{x\geq t\}}(x)+ 1_{\{x< t\}}(x) \frac{e^{\lambda x}}{e^{\lambda t}} \\
\end{align*}
and for $x<0$ we have that $P(S_{N_t} \leq x)=0$.
