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.
Last $2$ digits of powers of $7$:
$7^1 = 07$
$7^2 = 49$
$7^3 = 43$
$7^4 = 01$
..................................
$7^5 = 07$
$7^6 = 49$
$7^7 = 43$
$7^8 = 01$
So the last two digits repeat in cycles of $4$.
Hence $7^{81}$ has the last two digits same as that of $7^1$. So answer is : $07$
Best Answer
You get to use arithmetic modulo $100$ when all you care about is the last two digits. Also, since $\varphi(100)=40$ and $17$ is relatively prime to $100$, you get to do arithmetic modulo $40$ in exponents.
So you can first focus on $17^{17}$ modulo $40$: $$\begin{align} 17^{17}&\equiv17\cdot289^8&\mod40\\ &\equiv17\cdot9^8&\mod40\\ &\equiv17\cdot81^4&\mod40\\ &\equiv17\cdot1^4&\mod40\\ &\equiv17&\mod40\\ \end{align}$$
So we have that $$17^{17^{17}}\equiv17^{17}\mod100$$
$$\begin{align} 17^{17^{17}}&\equiv17^{17}&\mod100\\ &\equiv17\cdot289^8&\mod100\\ &\equiv17\cdot89^8&\mod100\\ &\equiv17\cdot(-11)^8&\mod100\\ &\equiv17\cdot121^4&\mod100\\ &\equiv17\cdot21^4&\mod100\\ &\equiv17\cdot441^2&\mod100\\ &\equiv17\cdot41^2&\mod100\\ &\equiv17\cdot(50-9)^2&\mod100\\ &\equiv17\cdot(2500-900+81)&\mod100\\ &\equiv17\cdot81&\mod100\\ &\equiv17\cdot(-19)&\mod100\\ &\equiv-\left(18^2-1\right)&\mod100\\ &\equiv-323&\mod100\\ &\equiv77&\mod100\\ \end{align}$$