5 residues mod America?

My approach is similar to Priyanshu's, but I will organize my work a bit differently.

Clearly, the given number $N$ is divisible by $5^5=3125$ .

To find the congruency class of $N$ modulo $2^5=3$ 2, we can work module $\phi(32)=16$ in the first exponent, modulo $\phi(16)=8$ in the second exponent, modulo 4 in the fourth exponent, and modulo 2 in the top exponent. Thus we have $N\equiv 5^{5^5}\equiv 5^5 \pmod {32}$

Since $N\equiv 5^5$ both modulo $5^5$ and modulo $2^5$ , the congruency holds modulo $2^5\times 5^5=10^5=10000$ as well, so that $N$ ends in $...03125$ , and the fifth digit from the end is $\boxed{0}$

Set $m$ $=$ $\huge\ { 5 }^{ { 5 }^{ { 5 }^{ { 5 }^{ 5 } } } }$ .

We have $\large\ { 5 }^{ \phi (2^{ 5 }) }\equiv 1 \pmod{ 2 }^{ 5 })$ .

Let us reduce $\large\ { 5 }^{ { 5 }^{ { 5 }^{ 5 } } } modulo { 2 }^{ 4 } = 16$ . Repeating the same trick, we want to reduce its exponent modulo $\phi (16) = 8$ . This amounts to computing $5^5$ modulo $4$ , and this is $1$ . Going backwards, we have the given number is congruent to $5^5 \pmod {32} = 5^5$ .

This shows that $m = { 5 }^{ 5 } + { 2 }^{ 5 }k$ , for some $k$ . This further shows that the residue of $m$ modulo ${ 10 }^{ 5 }$ is $3125$ . The fifth-to-last digit of $m$ is $\boxed{0}$ .