prime sequence prove that $p_n\ne 5$ i was preparing for olympiad then I got this question
Let $$<p_1,p_2,,,,p_n,,,>$$ be a sequence of primes defined by$p_1=2$ and for $n\ge 1$,$p_{n+1}$ is the largest prime factor of $p_1p_2...p_n+1$.Prove that $p_n\ne 5$ for any $n$.
I put $p_2=3,p_3=7$ 
I think sequence is increasing

Suppose $p_k=5$ for some $k>1$. This means that $p_1\cdots p_{k-1}+1=2^a3^b5^c$ for $c>0$. Since $p_1=2$ and $p_2=3$, $a=b=0$. So $p_1\cdots p_{k-1}+1=5^c$. Then:
$p_1\cdots p_k+1=(5^c-1)5+1=5^{c+1}-4.$
However,
$$5^{c+1}-4\equiv 2^{c+1}-1\pmod{3}\equiv 1 \pmod{3}.$$
So, $2^{c}\equiv 1\pmod{3}$. So $c$ must be even, e.g. $c=2d$. A quick calculation gives $p_3=7$. Yet, $5^{2d}-4\equiv(-3)^d+3\pmod{7}$, which is never equivalent to 1 mod 7. 
