# Pattern in power towers of 2 involving last digits

We have \begin{align} 2^{2^{2}} &\mod 10 = 6 \\ 2^{2^{2^2}} &\mod 100 = 36 \\ 2^{2^{2^{2^2}}} &\mod 1000 = 736 \\ 2^{2^{2^{2^{2^{2}}}}} &\mod 10000 = 8736 \\ 2^{2^{2^{2^{2^{2^2}}}}} &\mod 100000 = 48736 \end{align}

I think you get the point. Basically, it seems like $$^n2 \equiv ^{n+1}2 \mod 10^{n-2}$$ for $$n \geq 3$$, where $$^n 2$$ represents tetration. How could one go about proving this?

Let $$x_0=4$$ and $$x_{n+1}=2^{x_n}$$. The claim is $$x_{n+1}\equiv x_n\pmod{10^n}$$. By induction on $$n$$, we assume $$x_{n+1}\equiv x_n\pmod{10^n}$$ and we prove $$x_{n+2}\equiv x_{n+1}\pmod{10^{n+1}}$$.
We have $$x_{n+2}-x_{n+1}=2^{x_{n+1}}-2^{x_n}=2^{x_n}(2^{x_{n+1}-x_n}-1)$$ Since $$x_n\geq n+1$$ we get $$x_{n+2}\equiv x_{n+1}\pmod{2^{n+1}}$$. On the other hand for $$n\geq 2$$ we have $$x_{n+1}\equiv x_n\pmod{4}$$ and $$x_{n+1}\equiv x_n\pmod{5^n}$$ by assumption, so that $$x_{n+1}-x_n\equiv 0\pmod{4\cdot 5^n}$$. Since $$\varphi(5^{n+1})=4\cdot 5^n$$, this gives $$2^{x_{n+1}-x_n}\equiv 1\pmod{5^{n+1}}$$, thus giving $$x_{n+2}\equiv x_{n+1}\pmod{5^{n+1}}$$. This, together with $$x_{n+2}\equiv x_{n+1}\pmod{2^{n+1}}$$ gives $$x_{n+2}\equiv x_{n+1}\pmod{10^{n+1}}$$ concluding the proof.