The Agony of Extraordinarily Huge numbers

It's not hard, just tedious.

By Euler's theorem, we know that $a^{\varphi(n)} \equiv 1 \pmod{n}$ where $\varphi(n)$ is the Euler totient function and $gcd(a,n) = 1$ . Thus, equipped with this fact and that $\varphi(\varphi(100)) = \varphi(40) = 16$ we work from the inside out.

$15^{13^{11^{\ldots}}} \equiv (-1)^{13^{11^{\ldots}}} \equiv -1 \pmod{16}$

$17^{15^{13^{\ldots}}} \equiv 17^{-1} \equiv 33 \pmod{40}$

$19^{17^{15^{\ldots}}} \equiv (-3^{4})^{33} \equiv -3^{12} \equiv \boxed{59} \pmod{100}$

Let's see, and use good old modular arithmetic. This is a bit tedious too, but you can find that the last two digits in the powers of $19$ repeat themselves in intervals of $10$ . You can just do this by taking last two digits and then multiplying it over and over and over again with $19$ . The series is as given:

$1,19,61,59,21,99,81,39,41,79,1,...$

Now we just have to fund out the value of $n$ where $n\equiv17^{ 15^{ { \dots }^{ { 3 }^{ 1 } } } }\pmod{10}$

It's easy to observe that the units digits of the powers of $17$ repeat themselves at intervals of 4: $1,7,9,3,1,...$

Now we have to find what the last digit of this power of $17$ is, so that we can find $n$

$15^{ { \dots }^{ { 3 }^{ 1 } } }\equiv(-1)^{ { \dots }^{ { 3 }^{ 1 } } }\pmod{4}$

which means that $15^{ { \dots }^{ { 3 }^{ 1 } } }\equiv3\pmod{4}$

this implies that the power of $17$ had units digit $3$ and hence $n=3$ which gives the answer as $\boxed{59}$