\begin{align} X_1 &= 3 \\ X_2 &= 3^3 \\ X_3 &= 3^{3^3} \end{align}
Prove by induction that $X_n \equiv 3 \mod{4}$, for all $n \geq 1$.
My attempt at the induction step:
We assume that it holds for some arbitrary number $k \geq 1$, such that $X_{k} = {X_{k-1}}^3 \equiv 3 \mod{4}$. Thus, we have $X_{k} \equiv 3 \mod{4}$.
Now we need to show that if this is true for $k+1$. Thus, we have: \begin{align} X_{k+1} &= {{X_{k}}}^{3} \\ \log{X_{k+1}} &= 3\log{{X_{k}}} \\ \log{X_{k+1}} &\equiv 3 \log({3}\mod{4}) \\ \end{align} But I am not sure how to proceed as I do not know how to deal with the logarithm while still isolating the $X_k$. I would appreciate any guidence!
If $X_k\equiv3\pmod4$, then $X_k=4m+3$, for some $m\in\Bbb Z_+$. But then\begin{align}X_{k+1}&=3^{X_k}\\&=3^{4m+3}\\&=(3^4)^m\times3^3\\&=81^m\times3^3.\end{align}Now, use the fact that $81^m\equiv1\pmod4$ (since $80=20\times4+1$) and that $3^3\equiv3\pmod4$.