<s>[INST] The last 2 digits of $7^{7^{7^7}}$ What is the calculation way to find out the last $2$ digits of $7^{7^{7^7}}$? WolframAlpha shows  $...43$.
 [/INST] $7^2=49=50-1$
$\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod {100}$
Alternatively, $100=4\cdot25, 7^2\equiv1\pmod 4$ 
and $7^2\equiv-1\pmod{25}\implies 7^4\equiv(-1)^2\pmod{25}\equiv1$
$\implies 7^{\text{lcm}(2,4)}\equiv1\pmod {4\cdot25}$ as $(25,4)=1\implies 7^4\equiv1\pmod {100}$
So, we need to determine $7^{7^7}\pmod 4$
As $7^7$ is odd, and $7\equiv-1\pmod4\implies 7^{7^7}\equiv-1\pmod4\equiv3$
$\implies 7^{7^7}=4n+3$ for some integer $n>0$
So, $7^{7^{7^7}}=7^{4n+3}\equiv (7^4)^n\cdot7^3\pmod{100}\equiv1^n\cdot343$
</s>