Find the last two digits of $9^{{9}^{9}}$ [duplicate]

I have to find the last two decimal digits of the number $9^{{9}^{9}}$.

That's what I did:

$$m=100 , \phi(m)=40, a=9$$ $$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$

$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$

So,the last two digits are $8 \text{ and } 9$. $$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one?

Key remark: By the binomial theorem, for every odd $k$, $(10-1)^k=n+10\cdot k-1$ for some integer $n$ which is a multiple of $100$. Since $10\cdot k=10\cdot\ell(k)\bmod{100}$ where $\ell(k)$ denotes the last digit of $k$, this shows that, for every odd $k$, $9^k=10\cdot \ell(k)-1\bmod{100}$.

First application: $\ell(9)=9$, hence the key remark above yields $9^9=10\cdot \ell(9)-1=89\bmod{100}$.

Second application: our first application implies that $\ell(9^9)=9$ hence, using the key remark once again but this time for $k=9^9$, one gets $9^k=10\cdot\ell(k)-1=89\bmod{100}$.

And so on: for every tower of nines, $9^{9^{9^{9^{\cdots}}}}=89\bmod{100}$.

You could also just do modular exponentiation and take note of the periods.

The powers of $9 \mod 100$ are: 1, 9, 81, 29, 61, 49, 41, 69, 21, 89, 1, 9, 81, 29, 61, 49, ... They have a period of $10$. This means that if $n \equiv 9 \mod 10$, then $9^n \equiv 89 \mod 100$. Since $9^9 = 387,420,489$, this means that $9^{9^9} \equiv 89 \mod 100$.