If $f(x) =x^{x^{x^{x}}}$, determine the last two digits of $f(17) + f(18) + f(19) + f(20).$ If 
$$  f(x)= x^{x^{x^x}}, $$
then calculate the last two digits of $\displaystyle f(17) +  f(18) +  f(19) + f(20).$
I know that last two digits of $ f(20)$ will be $00$ and I also tried to find the last two digits by binomial coefficients,
like for $ 19^{19} ,$ we can get the last digit by : $$ {19 \choose 1} \times 20 - 1 $$ will be equal to $79$. Here I expresses $19$ as $20-1$ but for the next step I would need the value of $ 19^{19} $ and that would not be worth it !
Please help me, and tell me some good method to do  such type of questions.
Thank you in advance. Please help me out with the tags. I do not know under which section it should come.
 A: For example, let's consider $f(18)$.  It's easy to see $f(18) \equiv 0 (\mod 4)$.  What about mod $25$?  Well, $\phi(25) = 20$, so let's consider $18^{18^{18}} \mod 20$.
Again, $18^{18^{18}} \equiv 0 \mod 4$, so let's consider it mod $5$. $\phi(5) = 4$, and $18^{18} \equiv 0 \mod 4$, so $18^{18^{18}} \equiv 1 \mod 5$.  By Chinese Remainder,
$18^{18^{18}} \equiv 16 \mod 20$. So $f(18) \equiv 18^{16} \equiv 1 \mod 25$. Again by Chinese Remainder, since $f(18) \equiv 0 \mod 4$ and  $f(18) \equiv 1 \mod 25$, 
$f(18) \equiv 76 \mod 100$.
A: Since you seem to be having trouble following the hints given in the comments, I'll give a more extensive explanation.
In order to evaluate $$19^{19^{19^{19}}}\pmod{100}$$ we first want to find the smallest exponent $e$ such that $$19^e\equiv1\pmod{100}$$  Then if $n$ is a positive integer, we can write $n=me+k$ for integers $m$ and $k$, where $0\leq k<e$, so that 
$$19^n=19^{me}\cdot19^k=\left(19^e\right)^m\cdot19^k\equiv1^m\cdot19^k\equiv19^k\pmod{100}$$
We know $\phi(100)=40$, so by Euler's theorem, $19^{40}\equiv1\pmod{100}$, and the smallest exponent must divide $40$.  We find by experiment that the smallest exponent that works is $10.$  (Even though your calculator may not be able to compute $19^{10}$ directly, you only have to retain the last two digits as you compute higher powers, so this is not a problem.  You can shortcut the work by repeatedly squaring, and multiplying appropriate powers of $2$.) 
Anyway, now we need to know the value of $19^{19^{19}}$ modulo $10$.  We do the same thing as before, only we work modulo $10$ instead of modulo $100$ this time.  It's clear that $19^2\equiv1\pmod{10}$ so $2$ is the smallest exponent.  Now we know $19^{19}$ is odd, so $$19^{19^{19}}\equiv19^1\equiv9\pmod{10},$$ and 
$$19^{19^{19^{19}}}\equiv19^9\pmod{100}$$ 
If you've saved the powers of $19$ from when you figured out $19^{10}\equiv1\pmod{100}$, you can easily compute that $$19^9\equiv79\pmod{100}$$
When we try to apply this method to $18^{18^{18^{18}}}$, we run into the problem that Euler's theorem doesn't apply, since $18$ and $100$ are not relatively prime.  I see however that Robert Israel has just posted an answer explaining that case, so I'll stop here.   
