Calculate: $177^{20^{100500}}\pmod{60}$

The easiest way to handle such extremely powers is to calculate the power modulo $3$, $4$ and $5$ and apply the chinese remainder theorem. These modulo calculations are not very difficult :

Modulo $3$ is trivial; $177$ is divisble by $3$, so the residue is $0$.

Modulo $4$ is trivial as well ; $177$ has residue $1$ modulo $4$, so the power has residue $1$ modulo $4$ as well.

For Modulo $5$, you can reduce the exponent modulo $4$, which gives $0$, so the residue modulo $5$ is $2^0=1$.

So, the power is congruent $0$ modulo $3$ , $1$ modulo $4$ and $1$ modulo $5$. That gives $21$ modulo $60$

You got off to a good start there.

You know about Euler's Theroem, and Euler's totient, so I can add another tool to the box with the Carmichael function $\lambda$ which will give you the largest exponential cycle length (and still a value that all shorter cycles will divide). This combines prime power values through least common multiple rather than simple multiplication as for Euler's totient.

Here $\lambda(60) ={\rm lcm}(\lambda(2^2),\lambda(3),\lambda(5)) ={\rm lcm}(2,2,4) =4$. So for any odd number $a$, since there are no higher odd prime powers in $60$, you will have $a^{k+4}\equiv a^k \bmod 60$ for $k\ge 1$. (For even numbers you might need $k\ge 2$, since $2^2 \mid 60$). So $20^{100500}$ is just a huge multiple of $4$, and we can cast out all those $4$s all the way down to $3^4$. So the final result is

$$177^{20^{100500}} \equiv \underset {(\text{your result})}{3^{20^{100500}}}\equiv \underset {(\lambda(60)=4)}{3^4}\equiv 81\equiv 21 \bmod 60 $$

$3^5=3\pmod{60}$ so take the exponent modulo $4$ and then check to see if you need to multiply by $-1$.