Is the last digit of $2^{2^{n-1}(2^n-1)}-1$ always $5$ for $n >3$?

Question

Is the last digit of $2^{2^{n-1}(2^n-1)}-1$ always $5$ for $n >3$?

I did modification to the Mersenne numbers (Even perfect numbers) foruma I put that formual to be the power of 2 have got : $2^{2^{n-1}(2^n-1)}-1$ , Now for some values of $n >3$ that i have tried i have got the produced integer with last digit equal $5$ like this example, Now to do more $n$ or so large to conjecture that it is beyond of current tecknology , Really am interesting to prove that if it is true ?

Answer 1

We start from.the sequence: $$2\;\;4\;\;8\;\;16\;\;32\;\;64\;\;128\;\;256\;\;512\;\;1024\;\;2048\;\;4096$$ In particular, we are interested in the power of $2$ such that their last digit is $6$. This occours when $e\equiv0\mod 4$, in fact: $$2^e\equiv 6 \mod 10 \leftrightarrow e=4n, n\in N$$ Now, to show that $2^{2^{n-1}(2^n-1)}-1\equiv 5 \mod 10$, we need to show that: $$2^{n-1}(2^n-1)\equiv 0\mod 4$$ Let $n=3+k, k\in N_0$, we have: $$2^{3+k-1}(2^{3+k}-1)=2^2\cdot2^k\cdot(2^{3+k}-1)=4\cdot(2^{3+k}-1)$$ We are in the form $4a \mod 4$, where $a=(2^{n}-1)$, so: $$2^{n-1}(2^n-1)\equiv 0\mod 4, \forall n\in N, n>3$$

Note that this proprety holds also for $n=3$ because in that case, we can pick up a $2^2=4$ and so all the expression is congruent to $0$ modulo $4$.

Answer 2

Since $N=2^{2^{n-1}(2^n-1)}-1$ is odd, it is enough to show $5\mid N$, and the conclusion follows. But since $2^4 \equiv 1 \pmod 5$, we have $2^{2^{n-1}(2^n-1)}-1=(2^4)^{2^{n-3}(2^n-1)}-1\equiv 1-1=0\pmod {5}$ and we are done.