<s>[INST] To find the last digit of the number $4^{7^5}$? To find the last digit of the number $4^{7^5}$, I use the following method. 
first, we know $4\equiv_{10}4,4^2\equiv_{10}6,4^3\equiv_{10}4,...,4^{2n}\equiv_{10}6,4^{2n+1}\equiv_{10}4$
Then$7^1\equiv_21,7^2\equiv_21,7^3\equiv_21,...,7^5\equiv_21$
Therefore, $4^{7^5}\equiv4^{2k+1}\equiv_{10}4$.(because of the remainder 1)
and if to find the last digit of $4^{8^3}$ instead, the answer will be 6. Because $8^n\equiv_20,n\in \mathbb{N^+}$ so $8^n=2k,k\in\mathbb{N^+}$. From above, $4^{2k}\equiv_{10}6$
Is my method correct?
 [/INST] Your method is correct. Another way is to consider the easy fact that
$$2^{4n+k}= \begin{cases}2\pmod {10}\text{ for}\space k=1\\4\pmod {10}\text{ for}\space k=2\\8\pmod {10}\text{ for}\space k=3\\6\pmod {10}\text{ for}\space k=0\end{cases}$$ and the calculation $$7^5=16807$$ Then $$4^{7^{5}}=2^{33614}=2^{33612+2}=4\pmod{10}$$
</s>