Since $p_1p_2\cdots p_{n-1}$ is even, we know that $p_1p_2\cdots p_{n-1}+1$ is odd. This establishes that for all $p_n$ except when $n=2$, $p_n$ is odd. Moreover, it is easy to calculate that $p_2=3$. We will now proceed by contradiction. If some $p_n=5$, then $$p_1p_2\cdots p_{n-1}+1=2^x3^y5^z$$ for nonnegative $x,y,z$. But $p_1=2$ and $p_2=3$, so $x=y=0$. Therefore, $$p_1p_2\cdots p_{n-1}+1=5^z \implies p_1p_2\cdots p_{n-1} =5^z-1.$$ But, $5^z-1$ is divisible by $4$ and $p_1p_2\cdots p_{n-1}$ is not divisible by $4$, so we have a contradiction. Hence, $p_n\ne5$ for all $n$.