I have been given the following problem about a Poisson process with rate $\lambda$: show that $N_t$, the increments, and $\bar{S}=S_1+S_2+...+S_{N_t+1}$, $S_i$ the inter-arrival times, are independent random variables by determining $P(N_t=n ~\text{and}~ \bar{S}>t+s)$.
Here is what I have done: using the multiplication rule, $P(N_t=n \cap \bar{S}>t+s)=P(N_t=n | \bar{S}>t+s)P(\bar{S}>t+s)$. Given $N_t=n$ and $t=\sum_{i=1}^nS_i$, $\bar{S}=\sum_{i=1}^{n=N_t}S_i+S_{N_t+1}$. Then $P(\bar{S}>t+s)$ becomes $P(S_{N_t+1}>s)=e^{-\lambda s}$. I think $P(N_t=n | \bar{S}>t+s)=1$ resulting in $P(N_t=n \cap \bar{S}>t+s)=e^{-\lambda s}$.
However I do not see how this can help in proving that $N_t$ and $\bar{S}$ are independent with $P(N_t=n \cap \bar{S}>t+s)=P(N_t=n)P(\bar{S}>t+s)$?
$$P(N_t=n\wedge \bar S>t+s)=P(N_t=n\wedge N_{t+s}-N_t=0)=P(N_t=n)P(N_{t+s}-N_t=0)=P(N_t=n)P(N_s=0)$$
Then: $$P(\bar S>t+s)=\sum_{n=0}^{\infty}P(N_t=n\wedge \bar S>t+s)=\sum_{n=0}^{\infty}P(N_t=n)P(N_s=0)=P(N_s=0)$$
so that:$$P(N_t=n\wedge \bar S>t+s)=P(N_t=n)P(N_s=0)=P(N_t=n)P(\bar S>t+s)$$