5s over, now 9s!

We can use binomial theorem too.

${ 9 }^{ { 9 }^{ 9 } }\quad =\quad { (10-1) }^{ { 9 }^{ 9 } }$ ${ (10-1) }^{ { 9 }^{ 9 } }\quad =\quad { (-1+{ 9 }^{ 9 }\times 10+\frac { { 9 }^{ 9 }\times { (9 }^{ 9 }-1) }{ 2 } \times 100+...) }$

considering only the ten's and unit digit,

${ (-1+{ 9 }^{ 9 }\times 10+...) }$

$={ (-1+({ 10-1) }^{ 9 }\times 10+...) }$

$={ (-1+({ -1+9\times 10) }\times 10+...) }$

$={ (-1-10+900+...) }$

$=89+800+...$

Thus answer is $\boxed{ 89 }$

The last two digits of the powers of 9 are cyclical with an order of 10. This can easily be shown with a little number crunching (if there is a more elegant way to show this that I am unaware of please share!).

They are (in order and starting with a power of 0): {01, 09, 81, 29, 61, 49, 41, 69, 21, 89}

After this they start to repeat.

Therefore, if k is a specific power of 9, then k mod 10 will tell us the last two digits of $9^{k}$ .

Let $9^{9} = k$

The last 2 digits of $k$ are "89" since 9 mod 10 is 9, so $k$ mod 10 is also 9. We know this because the last digit of a number tells us what it's value mod 10 is.

This means $9^{k}$ has "89" as it's last two digits as well, so the answer is 89.

We use Euler's theorem to solve this problem. The answer we need is simply the residue of the given number modulo $100$ .

Denote by $\phi(n)$ the Euler's totient function. Now,

$\phi(100)=\phi(2^2\times 5^2)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=40$

Now, we have $\gcd(9,100)=1$ , so we're free to use Euler's Theorem to get,

$9^{9^9}\equiv 9^{9^9\bmod~40}\equiv 9^{\left((81)^4\cdot 9\right)\bmod~40}\equiv 9^9\equiv 29^3\pmod{100}$

Now, one can use one of two ways:

a) Simply calculate that value since it's fairly easy to compute $29^3$ and find the residue modulo $100$ . (boring)

b) Easily compute the residues modulo $4$ and $25$ and then combine the results using Chinese Remainder Theorem. (interesting)

Either way, we'll get $29^3\bmod 100=89$ and hence it follows that,

$\large 9^{9^9}\equiv 89\pmod{100}$

$By\quad Euler's\quad Theorem,\quad { 9 }^{ \varphi (x) }\equiv 1\quad (mod\quad 100)\quad where\quad \varphi (x)=40\\ Consider\quad { 9 }^{ 9 }\equiv 81\times 81\times 81\times 81\times 9\equiv 9\quad (mod\quad 40)\\ Therefore,\quad using\quad Euler's\quad Theorem,\\ { 9 }^{ { 9 }^{ 9 } }\equiv { 9 }^{ 9 }\quad (mod\quad 100)\quad \\ \quad \quad \equiv { 729 }^{ 3 }\quad (mod\quad 100)\\ \quad \quad \equiv { 29 }^{ 3 }\quad (mod\quad 100)\\ \quad \quad \equiv 24389\quad (mod\quad 100)\\ \quad \quad \equiv 89\quad (mod\quad 100)$