Towers of Powers of 17

By Euler's theorem we need to determine the value of each exponent below:

$\large\begin{array}{c}&\color{#20A900}{17}^{\color{#EC7300}{17}}&\equiv\left(16+1\right)^{17}&\equiv1\pmod{\color{#D61F06}{8}}\\ \color{#3D99F6}{17}^{\color{#20A900}{17}^{\color{#EC7300}{17}}}&\equiv\color{#3D99F6}{17}^{1}&\equiv1\pmod{\color{#D61F06}{16}}\\ \color{#D61F06}{17}^{\color{#3D99F6}{17}^{\color{#20A900}{17}^{\color{#EC7300}{17}}}}&\equiv\color{#D61F06}{17}^{1}&\equiv17\pmod{\color{#D61F06}{40}}\\ \large\color{#E81990}{17}^{\color{#D61F06}{17}^{\color{#3D99F6}{17}^{\color{#20A900}{17}^{\color{#EC7300}{17}}}}}&\equiv\color{#E81990}{17}^{17}&\equiv\left(21\right)^{4}\times{17}\equiv81\times17\equiv77\pmod{100}\end{array}$

NOTE:

$\color{#D61F06}{8}=\phi(16)=16\times\left(1-\frac{1}{2}\right)$

$\color{#D61F06}{16}=\phi(40)=40\times\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{5}\right)$

$\color{#D61F06}{40}=\phi(100)=100\times\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{5}\right)$

Checking for the last two digits of the $17^n$ , for $n \in [0,99]$ , shows that they repeat in a cycle of $20$ . And by induction, the pattern is true for all $n$ .

It can be seen that $17^n \equiv 17^{n \text{ mod 20}} \pmod{100}$ according to the table below:

We note that $17^{17} \equiv 77 \pmod{100}$ and since $77 \equiv 17 \pmod{20}$

$\Rightarrow 17^{17^{17^{17^{17}}}} \equiv 17^{17^{17^{77\pmod{100}}}} \equiv 17^{17^{77\pmod{100}}} \equiv 17^{77\pmod{100}}$

$\quad \quad \quad \quad \quad \equiv \boxed{77} \pmod{100}$