Inspired by Gavin Lo

Since $\lambda(100)=20$ and $\lambda(20)=4$ , where $\lambda(n)$ is the Carmichael Lambda, we can reduce the top exponent modulo 4, so that $9^{9^9}\equiv{9^{9^1}}=9^9\pmod{100}$ .

Now $9\times{9^9}=9^{10}=3^{20}=3^{\lambda(100)}\equiv{1}\pmod{100}$ . Multiplying both sides with $-11$ (the multiplicative inverse of $9$ ), we find $9^9\equiv{-11}\equiv\boxed{89}\pmod{100}.$

Note first that both $9$ and $9^{9}$ are odd.

Now by the binomial theorem, for any odd (positive) integer, we have that

$9^{n} = (10 - 1)^{n} = 100*C + 10n - 1$ for some $C \ge 0.$

(This is because $10^{2}$ will be a factor in every other term of the expansion.)

Now with $k$ being the units digit of $n,$ we have that $10n \equiv 10k \pmod{100}.$

Thus, for odd $n$ , $9^{n} \equiv (10k - 1) \pmod{100},$ and so $9^{9} \equiv 89 \pmod{100}.$

But then, as the last digit of $9^{9}$ is $9,$ we in turn have that $9^{9^{9}} \equiv \boxed{89} \pmod{100}.$

As one can see, the same result will hold for any "power tower" of $9$ 's.

A simple and hit-and-trial method... worked for me.... 9^n ends with 1 for even n and with 9 for odd n... this leaves us with two options 09 and 89... On careful observation.. it can be observed that after 9^1 ending with 09... 9^n ends with 09 next time with n=11... therefore for two digits... the pattern repeats with every 10 numbers..... also 9^9 ends with 89... and 89 ends with 9... therefore... (9^9)^9 must end with 89.... Sorry... but at present level... I don't know the concepts of mod.... so I didn't use it....

As you can see the concept, if we raised 9 by an odd no the last digit is 9, and based on the choices 89 is correct since second digt must not be zero.