Proof involving the infinite number of primes 
Given that
$R = p_1p_2\cdots p_n + 1$ where $p_1 < p_2 < \cdots  < p_n$ and $p$ are prime numbers.
Prove that if $R$ is not prime then $R$ must have a prime factor $q$ that is larger than $p_n$.

I directly understand that this question refers to Euclid's infinite primes proof however; Honestly, I don't know really how to even begin this problem.
Any advice on this problem would be very helpful and appreciated.
 A: It's not true: If $p_1=5$ and $p_2 = 7$, and that's your sequence, then $p_1 p_2 + 1$ has prime factors $2$ and $3$.  Those numbers are not bigger than $5$ and $7$.
What Euclid showed is that if you have any finite set of prime numbers, and you multiply them and then add $1$, then the prime factors of the number you get are not in the finite set you started with.  That's not the same as saying they're bigger than any of the numbers in the set you started with.  That's not true: sometimes they're actually smaller than all of them.
So you have some primes $p_1,\ldots,p_n$.  The number $R=(p_1\cdots p_n)+1$ leaves a remainder of $1$ when divided by any of $p_1,\ldots,p_n$; therefore it is divisible by none of them.  Its prime factors are therefore not in the set $\{p_1,\ldots,p_n\}$.
For example, consider
\begin{align}
888 & = 2\cdot3\cdot37 \\
889 & = 7\cdot127 \\
890 & = 2\cdot5\cdot89 \\
891 & = 3\cdot3\cdot3\cdot3\cdot11 \\
    & {}\  \vdots
\end{align}
The next number after $888$ cannot be $37$ because you don't get another number divisible by $37$ until you've gone $37$ steps beyond that, up to $888 + 37$.  It cannot be divisible by $3$ since you don't get another number divisible by $3$ until you get to $888+3$.  It cannot be divisible by $2$ since you don't get another number divisible by $3$ until you get to $888+2$.  Therefore its prime factors are not in the set $\{2,3,37\}$.
