Prime tower

We have $\lambda(100)=20$ and $\lambda(20)=4$ , where $\lambda(n)$ represents the Carmichael Lambda . Thus $13^{13^{13}}\equiv 13^{13} \pmod{100}$ since $13\equiv 1 \pmod{4}$

We observe that $13^3\equiv -3 \pmod{100}$ so $13^{12}\equiv 3^4=81$ and $13^{13}\equiv 13\times 81\equiv \boxed{53}$

$13^{13^{13}} \pmod {100} \\ 13^{\phi(100)}=13^{40} \equiv 1 \pmod {100} \\ \Rightarrow 13^{13} \pmod {40} \\ (13^2)^6\cdot 13 \equiv (169)^{6}\cdot 13 \pmod {40} \\ 9^6\cdot 13 = 81^2\cdot 13\equiv 13 \pmod {40} \\ \therefore 13^{13} \pmod {100} \\ (13^3)^4\cdot 13 =(2197)^4\cdot 13 \pmod {100} \\ (-3)^4\cdot 13=81\cdot 13\equiv \boxed{53} \pmod {100}$

Slightly different method : Other than Euler's Theorem

For the tower , $13^{13^{13}}$

Let upper $13^{13} \equiv x \mod y$ , so $13^{13}=yz+x$ for some $z$ .

From here we get , $13^{yz+x} \mod 100$ .

Let us assume that , $13^y \equiv 1 \mod 100$ .

hence , $13^{yz} × 13^x \equiv 13^x \mod 100$

So , $13^y=100p+1$ for some $k$ . Note that numbet $13^y$ should end with 1 , which is only possible

when $y=4a$ for some $a$ .

So we arrive at , $13^{4a . z}\mod 100$ therefore $61^a\equiv 1 \mod 100$ , it is clear that least value of $a$ should be equal to 4

So $y=20$ .

So , $13^{13}\equiv x \mod 20$ hence $x=13$ .

Now the final congruence leads to , $13^{13} \mod 100$

and hence , $13^{13^{13}} \equiv 53 \mod 100$

Using Euler's Theorem we have $13^{ 40 }\equiv 1 (\mod{100})$

And since $13^{ 13 }\equiv 13 (\mod{40})$ , we have $13^{ { 13 }^{ 13 } }=13^{ 40x+13 }\equiv 1\cdot { 13 }^{ 13 }\equiv 53 (\mod{100})$

$3^m~ is~ cyclic,~ period~ 4.~ m\ \equiv~ n~ mod~ 4.~~ Then~ for~ ~n~=~1,~ 2,~ 3,~ 4,~ last~ digit~ of~ 3^m ~= ~3,~ 9, ~7,~ 1.\\ m=13, ~so~ n~= 3.\\ \text{Last two digit will complete a cycle when they are 01.}\\ \text{ So end of every 4th (n=4) term starting from exponent 1 is investigated.}\\ 13^4\ \equiv~{\Large 61} ~ mod~ 100, \\ 13^8\ \equiv~ 61*61\ \equiv~{\Large 21} ~ mod~ 100, \\ ~~~~The~ tenth~ digit~ is ~increased~ by~6.\\ \therefore~21+60=81,~\implies~13^{12}\ \equiv~{\Large 81} ~ mod~ 100, \\ \therefore~\color{#E81990}{13^{13}=13^{12}*13=81*13 \equiv~{\Large 53} ~ mod~ 100,} \\ \therefore~81+60=141,~\implies~13^{16}\ \equiv~{\Large 41} ~ mod~ 100, \\ \therefore~41+60=101,~\implies~13^{20}\ \equiv~{\Huge 01} ~mod~ 100, \\ \implies~The~cycle~for~last~two~digit~is~~20. \\ \therefore\ \Large 13^{13^{13}}\ \equiv\ 13^{53}\ \equiv\ 13^{{53}\ mod\ 20} \equiv~13^{13}\ \equiv~{\Huge 53} ~ mod~ 100.\\ Below~same~is~worked~out~in~tabular~form. \\$ $\color{#20A900} {13^1 \equiv~13 ~ mod~ 100\ \ \ \ \ \ \ \ 13^2 \equiv~69 ~ mod~ 100 \ \ \ \ \ \ \ \ 13^3 \equiv~97 ~ mod~ 100\ \ \ \ \ \ 13^4\ \equiv~61 ~ mod~ 100 \\ 13^5 \equiv~93 ~ mod~ 100\ \ \ \ \ \ \ \ 13^6 \equiv~09 ~ mod~ 100 \ \ \ \ \ \ \ \ 13^7 \equiv~17 ~ mod~ 100\ \ \ \ \ \ 13^8\ \equiv~21 ~ mod~ 100 } \\ ................................................................................ .... (++) \\ {\color{#D61F06}{13^{13} \equiv~53 ~ mod~ 100 } } ................................................................... (++++) \\ .................................................................. 13^{\color{#20A900}{20} } \equiv~01 ~ mod~ 100 \\ \therefore\ \Large 13^{13^{13}}\ \equiv\ 13^{{53}\ \ \equiv\ 13\ mod\ 20} \equiv~13^{13}\ \equiv~{\Huge 53} ~ mod~ 100.$

Note:-
$(++)\color{#E81990}{ \text{ 1st column tenth digit increment is 8, for 4th, it is 6.} \\ \text{ So for exponent 13, 1st column unit digit is 3. last one of tenth digit,=(1+3*8)=15 }\\ (++++) \text{So for last digits to be 01, with tenth digit increment of 6, exponent is 20.} }\\$
You may refer to my note in the solution of hard problem " Mod-Ladder" by Chinmay Sangawadekar for the logic of my above solution. The method can solve TOWER problem with the help only of a simple scientific calculator and no knowledge of Theory of Numbers.

I used the fact that ${(2m + 1)^{20n}}\equiv{01}\pmod{100}$ , where $n \in N , m \in N_0 , 5\nmid 2m + 1$

By Euler's Theorem , we have $13^{\phi(20)}\equiv{13^{12}}\equiv{1}\pmod{20}$

$\Rightarrow 13^{13}\equiv13^{12}\times{13}\equiv1\times{13}\equiv13\pmod{20}$

$\therefore 13^{13^{13}}\equiv {13^{(20n + 13)}}\equiv{13^{20n}\times{13^{13}}}\equiv{13^{13}}\pmod{100}$ .

$13^{13}\equiv(13^{3})^{4}\times{13}\equiv(2197)^{4}\times{13}\equiv(-3)^{4}\times{13}\equiv{81}\times{13}\equiv{53}\pmod{100}$

Our answer is $\boxed{53}$