TITLE: Method of calculating the last digit of $7^{49}$
QUESTION [6 upvotes]: I would like to point out the difference between $7^{7^7}$ and $7^{49}$. typically the notation $a^{a^{a^a}}$ means $a^{(a^a)}$. Heuristically, start from the top and work your way down. e.g. $2^{2^{{2^2}}} = 2^{(2^{({2^2})})} = 2^{(2^{4})} = 2^{16}=65536$.

REPLY [2 votes]: notice that $7^2 \equiv 49 \equiv -1 \pmod {10}$
then we have 
$$7^{49} \equiv 7^{32} \times 7^{16} \times 7 \equiv (7^{2})^{16} \times (7^{2})^{8} \times 7 \equiv(-1)^{16} \times (-1)^8 \times 7\equiv 7 \pmod {10}$$
thus the answer is 7.