$(n+1)$th prime $p_{n+1}$ less than or equal to $p_1p_2\dots p_n+1$
Solution 1:
To see that $p_{n+1}\le p_1p_2\dots p_n+1$, note that $p_1p_2\dots p_n+1$ must have a prime factor, but it cannot be any of $p_1, p_2, \dots, p_n$ by the same argument as in the proof. So any prime factor of $p_1p_2\dots p_n+1$ must be a new prime number less than or equal to $p_1p_2\dots p_n+1$. Therefore, the smallest prime larger than $p_n$ can be at most $p_1p_2\dots p_n+1$.
As for why $2^1 + 2^2 + \dots + 2^n = 2^{n+1}-2$, write it in binary.