Is this proof about Mersenne numbers acceptable? I want to prove following property of Mersenne numbers :

If $p > 3$ then $M_p\equiv 1 \pmod {6\cdot p}$

So, according to Fermat's Little Theorem we may write :
$2^p\equiv 2 \pmod p \Rightarrow 2^p-2=a\cdot p \Rightarrow 2^p-1=a\cdot p +1 \Rightarrow$
$\Rightarrow 2^p-1\equiv 1 \pmod p \Rightarrow M_p\equiv 1 \pmod p$
Now, since $M_p\equiv 7 \pmod {24} $ it follows that :
$M_p=24\cdot a +7 =24\cdot a+6+1=6\cdot(4a+1)+1 \Rightarrow M_p\equiv 1 \pmod 6$
Therefore we may conclude :
$(6 \mid (M_p-1) \land p \mid (M_p-1))\Rightarrow 6\cdot p \mid(M_p-1) \Rightarrow M_p \equiv 1 \pmod{6\cdot p}$
Am I correct ?

Note that $M_p= 2^p -1 = (2-1)(2^{p-1} + 2^{p-2} + \dots + 2^2 + 2^1 + 1)$.  If $p>3$ then modding out by $2^3=8$ leaves $2^2+2^1+1 =7$ behind.  Also $M_p = 2^p-1$ and $p$ is odd, so $2^p \equiv 2 \mod 3$, which means $M_p \equiv 1\mod 3$.  The Chinese Remainder Theorem then applies and shows that $M_p \equiv 7 \mod (3*8)$.
