How do I solve $13^{({11}^{7})}\equiv x\pmod{10}$ 
Solve $x$, where $x\in[0,9]\cap\mathbb{Z}:$
$$13^{11^7}\equiv x\pmod{10}$$

Here is my thoughts:
Since $13\equiv3\pmod{10}$
Implies $13^{11^7}\equiv 3^{11^7}\pmod{10}$
$$3^0=1,3^1=3,3^2=9,3^3=27,3^4=81\cdots$$
For mod $10$, have reminder repeat between:
(they have to repeat, but i will skip the induction here) $$R_1=\{1,3,9,7\}$$
Then solve
$$11^7+1\equiv x\pmod{|R_1|}$$
$$11^0+1=2,11^1+1=12,11^2+1=122,11^3+1=1332$$
For mod $4$, have reminder index repeat between:
$$R_2=\{2,0\}$$
In another word, the reminder repeat between second and last term in $R_1$
Then solve $$7+1\equiv x\pmod{|R_2|},\text{ that }x=0$$
$x=0$, means the reminder is the last term in $R_1$ which is $7$, that implies
$$13^{11^7}\equiv7\pmod{10}$$
Therefore $x=7$

My qeustion: is there any theorem to apply so this calculation could be simpler$?$
Any help would be appreciated.

HINT:
Using Euler's theorem,
$\phi(10) = 4 \Rightarrow 13^4\equiv 1 \pmod{10}$
and
$11^7 \equiv (-1)^7\pmod{4}$
