Why Do Mersenne Primes Only Occur at Terms of Prime Index? A Mersenne prime is a prime of the form $2^n-1$.
Only when $n$ is a prime itself is there a chance that $2^n-1$ is a Mersenne primes. The largest primes discovered are almost always Mersenne primes. Some of the more known Mersenne primes are $3, 7, 31, 127$, e.t.c.
Now on to the question.  

Why do non-prime values of $n$ never yield a prime?   

I have always heard from my teachers that Mersenne primes occur at only the prime values of $n$ but no one ever explained it to me. Is there any way of proving this?  Or are there any exceptions for $n>1$?  
P.S. As you may have guessed from my writing "teachers" instead of "professors", I am only in grade $10$ and not that skilled so I would prefer if you could give me simple explanations. Thanks in advance!
 A: Because if $m=kl$, with $k,l>1$, then\begin{align}2^n-1&=2^{kl}-1\\&=(2^k)^l-1^l\\&=(2^k-1)\bigl((2^k)^{l-1}+(2^k)^{l-2}+\cdots+1\bigr).\end{align}
A: I think that this is easiest seen when numbers are represented in binary form. A number $2^n - 1$ is the represented as $n$ ones, e.g. $255 = 2^8 - 1 = 11111111_2.$ When $n$ is not a prime, this number can be grouped into subparts, e.g. $1111~1111_2$ or $11~11~11~11_2$, and then easily be written as a product: $1111_2 \times 1~0001_2$ or $11_2 \times 1~01~01~01_2.$
A: That is because of theory pertaining to the family of Lucas sequences, which the sequence of Mersenne numbers is a member of. in particular, and applied here to the sequence of Mersenne numbers, $M_{n},$:
$$ M_{r} | M_{s} \,\,\,\, if \,\, and \,\, only \,\, if \,\,\,\, r | s.$$ 
Hence, if the index of some Mersenne prime was composite, say $mn$, then this would lead to a contradiction because $M_{mn}$ would then be divisible by both $M_{n}$ and $M_{m}$ contrary to the assumption that it is prime.  
