How to approach this sequence? (elementary number theory) Given is the following sequence: $a_1 = 1$ and $a_n$ equals the biggest prime divisor of $1+ a_1*\dots*a_{n-1}$ . 
It is then claimed: $11$ does not occur in this sequence.
How can one approach this problem? I first thought about trying a proof by contradiction. 
Let's then be $a_m$ the first occurrence of $11$ in this sequence. Then there exist $a,b,c,d, e \in \mathbb{N_0}$ such that:
$2^a * 3^b * 5^c * 7^d * 11^e = 1+ a_1*\dots*a_{m-1}$ .
From there however, it looks me like a dead end. It seems like I am missing a key observation. I would be happy, if anyone with more knowledge could offer an advice/suggestion how to begin here.
EDIT: Related: Euclid Mullin Sequence
 A: Note: Personally, I find my solution too short and "too easy" to be a solution to this problem. Therefore, feel free to make a note or comment, when there is an error.
I would "solve" this specific problem stated above like this: 
The first elements of the sequence can be computed fairly quickly:
$a_1 = 1, a_2 = 2, a_3 = 7, a_4 = 43, a_5 = 139$
Now following lulu's advice, we show that a prime number occurs at most twice in this sequence.
Given that $a_n = 1+a_1 * \dots * a_{n-1}$  it follows that $a_n \equiv 1\pmod {a_i} $ for $1 \le i \le n-1$, thus there is no $a_i$ with $a_i| a_n$ $\Rightarrow a_i \neq a_n 
 \ \forall i$ with: $ \ 1\le i \le n-1$
Now: Let $a_n = 11$ and $b_n = 1+ a_1* \dots \ * a_{n-1}$ then by definition $11|b_n$ and if $k|b_n$ and $k \neq 11$, then $k = 1 \lor k= 5$, in particular: $2,3,7 \nmid b_n$
$\Rightarrow b_n \le 55$, but: $\frac{b_n}{55} \ge \frac{a_1 * \dots * a_5 +1}{55} \gt 1 \Rightarrow$ $b_n$ has at least one prime factor greater than $11$, thus: $11$ can't occur in this sequence.
EDIT: Proving that does not occur in the sequence can be doen quickly:
If 5 occurred, then we had for some $b_n = 2^a * 3^b * 5^c = 1+2* \dots * a_{n-1} $. 
$a,b$ must be equal to $0$. If they weren't, then 
$2^a * 3^b * 5^c \equiv 0 \pmod {2,3} $, but $b_n \equiv 1 \pmod {2,3} $
Thus $b_n$ would be of the form: $b_n = 5^c$ for some $c$ however: 
$5^c - 1 \equiv 0 \pmod 4 \forall \ c \in \mathbb{N}$  and $a_1* \dots \ * a_{n-1} \equiv 2 \pmod 4 $ 
Thus $5$ cannot occur in the sequence. 
However there is still some way to go to prove $11$ does not occur and I actually don't see how one would do that. The same argument/trick, used for $5$, doesn't seem work.  
EDIT2 : $5^a * 11 ^b \equiv 3 \pmod 4$ as $2 | a_1 * \dots* a_{n-1} $
$\Rightarrow 1^a * 3^b \equiv 3 \pmod 4 \Rightarrow b$ is odd.
$3 | 5^a * 11^b -1 \Rightarrow (-1)^a *(-11)^b \equiv (-1)^{a+b} -1 \equiv 0 \pmod 3 \Rightarrow a+b $ even$ \Rightarrow a$ is also odd.
Then: $7 | 5^a * 11^b -1 \Rightarrow (-2)^a * 4^b \equiv (-2)^{a+2*b} \equiv 1 \pmod 7 $
This however is only the case: if $a+2*b$ is a multiple of $6$ (Little Fermat), which can't be the case. Hence: $11$ cannot occur in the series and we are done. Quite a journey over the past few days, but I believe the proof is now complete.
