Congruence involving CRT

Question

I was working on a problem, I arrived at the point at which I have to find $17^{{{17}^{17}}^{17}} \pmod {25}$

My attempt: $$ 17^{{{17}^{17}}^{17}}\equiv 17^{{{{17}^{17}}^{17}} \pmod{\phi(25)}} \pmod {25} $$$$17^{{{17}^{17}}}\equiv 17^{{{{17}^{17}}} \pmod{\phi(20)}} \pmod{20}$$$$17^{17}\equiv1^{17}\equiv1\pmod{\phi(20)=8} $$ Thus:$$17^{{{17}^{17}}^{17}}\equiv 17^{17}\equiv17^{-8}\equiv(17^{-1})^{8}\pmod{25}$$ The inverse of $17$ modulo $25$ its $3$ since $17\cdot3=51$, so:$$17^{{{17}^{17}}^{17}}\equiv3^{8} \equiv3^3\cdot3^3\cdot3^2\equiv2\cdot2\cdot9\equiv36\equiv11\pmod{25}$$ I checked but the solutions says that it is actually congruent to $2$ and not $11$, everything before the inverse it's fine for sure since OP follows the same procedure, what did I do wrong?

Answer 1

There is a mistake. $$17^{{{17}^{17}}^{17}}\equiv 17^{17}\equiv17^{\color{red}{-8}}\equiv(17^{-1})^{\color{red}{8}}\pmod{25}$$ should be $$17^{{{17}^{17}}^{17}}\equiv 17^{17}\equiv17^{\color{red}{-3}}\equiv(17^{-1})^{\color{Red}{3}}\pmod{25}$$