Infinitely many primes proof using Euler's totient I need something explained or corrected: 
In my number theory class we proved that there are an infinite number of primes using Euler's Phi Totient. It went something like this:
Let $M = p_1 p_2 \dots p_n$ be the product of all primes. Consider $1 < A \le M$:
Some prime must divide $A$, call it $q$. Since $q$ must be one of the primes, $q$ must divide $M.$
So the $gcd(A,M) > 1.$ Thus $\phi(M) = 1$ ...? 
Which is not even and contradicts the theorem that $\phi(N)$ is even for $N>2.$ Therefore there exists an infinite number of primes.
I get confused by the statement "Thus $\phi(M) = 1$"....
Did i possibly copy this proof down wrong? or am I missing something?
Thank you in advance. 
Edit: by consider a such that...i meant consider an integer '$a$' I replaced it with $A$ to hopefully make it more clear. 
Edit2: I'm sorry I am not familiar with using the equation editor. This is not a homework assignment, just studying for my exam. I just want to be able to understand this or clearify it. 
 A: By definition, $\phi(M)$ is the number of numbers in $S=\{1,2,\dots,M\}$ that are relatively prime with $M$. The argument shows that if $1<A\le M$, then $A$ is not relatively prime with $M$, so there is only one element of $S$ relatively prime with $M$, namely 1, and $\phi(M)$ is therefore equal to 1.
A: Note that $\phi(M)$ counts the number of integers in the interval $[1, M-1]$ which are relatively prime to $M$.  We are told to use the fact that $\phi(M)$ is almost always even.  
Since $1$ is relatively prime to $M$, that leaves $\phi(M)-1$ numbers in our interval, different from $1$, which are relatively prime to $M$.
But since $\phi(M)$ is even, it follows that $\phi(M)-1$ is odd, and in particular not equal to $0$, since $0$ is even!  Thus there is a number $a \ne 1$, in our interval, such that $a$ is relatively prime to $M$. Any prime divisor $p$ of $a$ must be different from all the $p_i$ in the given list, since $a$ is relatively prime to $M$.
Seems like a bit too much machinery for this problem, specially since we can see that if $n>1$, the number $M-1$ is not equal to $1$, and is relatively prime to $M$.
