All mersenne primes greater than 3 will end with either 1 or 7?

[soumya]

All mersenne primes greater than 3 will end with either 1 or 7.
Q1. Is the above statement true?
Q2. Why is it so?

[Batalov]

Q1. Trivially, yes.
Q2. Because for any odd n, 2^n = 2 or 8 (mod 10):
2^1 = 2 (mod 10)
2^3 = 4*(previous value) = 8 (mod 10)
2^5 = 4*(previous value) = 2 (mod 10)
2^7 = 4*(previous value) = 8 (mod 10)
2^9 = 4*(previous value) = 2 (mod 10), etc, 2 and 8 will alternate.
Now, with the exception of 3 (which is 2²-1), all other Mersenne primes are 2^{odd n} -1.

[Raman]

Originally Posted by soumya

All mersenne primes greater than 3 will end with either 1 or 7.
Q1. Is the above statement true?
Q2. Why is it so?

@soumya: Are you being aware that in addition to this thing,
and then stuff matters,
every prime factor for the Mersenne number with an odd exponent into this,
i.e. a number of the form 2^2x-1-1, x ≥ 2, 2x - 1 ≥ 3, it is being of the form congruent to 1 or 7 (mod 8)?

In addition to it, each prime factor for the part afterwards
after algebraic factors being removed for the Mersenne number 2^x-1, are always being of the form 2kx+1, i.e. it is being congruent to 1 (mod 2x).

and then certainly that always that
and then therefore, so, thus, that way,
it is being true enough statement for the Cunningham numbers b^x±1, as well as,

as follows as

Why so, thus?

and then

also too