I try to solve this theorem:
Theorem: Let $x_1, x_2,...$ iid with exponential distribution with rate $\lambda$. The density function of $S_n$ is given by: $$ P(S_n \le t)=\sum_{k=n}^{\infty}\frac{e^{-\lambda t}(\lambda t)^k }{k!}, t\ge 0. $$
I found one proof is by induction on n, but I'm not sure:
When $n=1$ $S_1$ consider that in a Poisson process, if events occur on average $\lambda t$ occurrences per $t$ units of time. The Poisson distribution describing this process is therefore:
$$P(S_1 \le t)=\frac{e^{-\lambda t}(\lambda t)^1 }{1!}=e^{-\lambda t}\lambda t $$
To do the induction step suppose that the formula is true for n. $$ P(S_n \le t)=\sum_{k=n}^{\infty}\frac{e^{-\lambda t}(\lambda t)^k }{k!}, t\ge 0. $$
So for $n+1$: $$ P(S_{n+1}<t_2)=P(S_{n}+X_{n+1}<t_2)=P(S_n<t)P(X_{n+1}<t_2-t) $$ $$ P(S_n<T)P(X_{n+1}<t_2-t)=\sum_{k=n}^{\infty}\frac{e^{-\lambda t}(\lambda t)^k }{k!}*\frac{e^{-\lambda t}(\lambda t)^{n+1} }{(n+1)!} $$
$$ \sum_{k=n+1}^{\infty}\frac{e^{-\lambda (t+t_2-t)}(\lambda (t+t_2-t))^k }{k!}=\sum_{k=n+1}^{\infty}\frac{e^{-\lambda t_2}(\lambda t_2)^k }{k!} $$ which completes the proof.
Based on Poisson process theory, the correct expression for $P[S_n \leq t]$ should be given by: $$P[S_n \leq t] = P[N(t) \geq n] = \sum_{k = n}^\infty e^{-\lambda t} \frac{(\lambda t)^k}{k!}.$$
So maybe what you tried to express is that $$P[S_n \color{red}{>} t] = 1 - P[S_n \leq t] = \sum_{k = 0}^{n - 1} e^{-\lambda t} \frac{(\lambda t)^k}{k!}.$$