Laplacian Rule of Succession. How to find $P(X_{n+1}=1 \mid S_n=n)$?

Question

Been working through stat110 and "https://youtu.be/N8O6zd6vTZ8?t=2990" talks about the laplcian rule of succession and I am not too sure how he got to his final step(49:50). The problem is as follows

Let $X_{1},X_{2},...,X{n}$ be i.i.d $bern~p$ if $p$ is known. We assume that $p$ ~ $unif(0,1)$. Let $S_{n}=X_{1}+X_{2}+...+X{n}$. So, we want to find how $p$ is distributed given that $Sn=n$. We also want to find $P(X_{n+1}=1 \mid S_{n}=n)$

In the video, the professor shows that the PDF of $p \mid S_{n}=n$ is $(n+1)p^n$

I understand what he is doing up till point but for the final part, if im understanding him correctly. He says that by the fundamental bridge $P(X_{n+1}=1 \mid S_{n}=n)$ = $E(X_{n+1} \mid S_{n}=n)$ and that that $X_{n+1} \mid S_{n}=n$ has the PDF we just found $(n+1)p^n$ and thus we can just take the integral $(\int_{0}^{1} (n+1)p^n \,dp)$

My confusion is as follow:

1. I am slightly confused about how he was able to use the fundamental bridge since our random variable isnt bern(p) where p is constant but instead where p is a variable.
2. It feels intuitive but I have no idea how to show that $X_{n+1} \mid S_{n}=n$ has the PDF we just found $(n+1)p^n$
3. 1. and 2. seem slightly contradictory since if 2. were true, then $X_{n+1} \mid S_{n}=n$ definitely wouldnt be bern(p) so how are we able to use the fundamental bridge