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.
