The last two digits of $9^{9^9}$

Euler's Theorem is not needed. It can be completely solved using only the Binomial Theorem:

$$\rm 9^{\color{#c00}{\large 10}} =\ (-1+10)^{\color{#c00}{\large 10}} =\: (-1)^{\color{#c00}{\large 10}} - \color{#c00}{10}\cdot 10 + 10^{\large 2}\:(\cdots)\ \color{}{\equiv\ 1}\ \ (mod\ 100)$$

So $\rm \bmod 100\!:\, \ 9^{\large 9^{\LARGE 9}}\!\!\equiv\ 9^{\large 9^{\LARGE 9}\, mod\ \color{#c00}{10}} \equiv\ 9^{\large (-1)^{\LARGE 9}}\!\! \equiv 9^{\large -1}\!\equiv \dfrac{1}9 \equiv \dfrac{-99}9 \equiv {-}11 \equiv 89 $

Remark $ $ Above we used the useful fact that if the powers of $\,a=9\,$ repeat with period length $\color{#c00}{10}\,$ then all exponents on $\,a\,$ can be taken modulo $\,\color{#c00}{10}.\,$ Said more precisely we used the following

$$\ \ \color{#c00}{a^{\color{#c00}{\large 10}}\equiv 1}\!\!\pmod{\!m},\,\ J\equiv K\!\!\!\pmod{\!\color{#c00}{10}}\ \,\Rightarrow\,\ a^{\large J}\equiv a^{\large K}\!\!\!\!\pmod{\!m}$$

for the specific values $\ a=9,\,$ and $\,J = 9^{\large 9},\,$ and $\,K = (9^{\large 9}\,{\rm mod}\ 10).\,$ A proof is easy:

$$ J = K\! +\! 10N\,\Rightarrow\, a^{\large J}\! = a^{\large K+10N}\! = a^{\large K} (\color{#c00}{\large a^{10}})^{\large N}\!\equiv a^{\large K} \color{#c00}1^{\large N}\!\equiv a^{\large K}\!\!\!\!\pmod{\!m}\qquad $$

where we have employed the $ $ Congruence Product and Power Rules. For further discussion see modular order reduction.

Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

By the binomial theorem, we have $$(-1+10)^9\equiv{9\choose0}(-1)^910^0+{9\choose1}(-1)^810^1+{9\choose2}(-1)^7{\color{Red}{10^2}}+\cdots$$ $$\equiv-1+90=89\pmod{10^2}.$$ (All summands with powers of $10$ greater than $1$, the first instance in red, can be ignored.)