Can't use Euler's Theorem? I want to use Euler's theorem to calculate the following:
$14^{(2019^{2019})} \mod 60$
$a^{\phi(n)} \equiv 1 \mod n \iff \gcd(a,n) = 1$
Now I start with the outer one first:
$14^{(2019)}\mod 60 \iff \gcd(14,60) = 2$
Now I don't know how can I solve it ...
Can anyone suggest me any hints/further calculations?
Thank you in advance,
Kind Regards.

Like How to find last two digits of $2^{2016}$, 
Get the last two digits of $16^{100}$ and $17^{100}$
what are the last two digits of $2016^{2017}$?
last two digits of $14^{5532}$?,
As $(14^n,60)=2^2$ for $n\ge2$
and as  $14\equiv-1\pmod{15}\implies14^m\equiv(-1)^m$
$\implies F_{m+2}=14^{m+2}=14^2\cdot14^m\equiv14^2(-1)^m\pmod{15\cdot14^2}$
$F_{m+2}\equiv14^2(-1)^m\pmod{15\cdot4}$ as $15\cdot4$ divides $15\cdot14^2$
$F_{m+2}\equiv16(-1)^m\pmod{60}$
If $m$ is odd like here $2019^{2019}-2,$ $$F_{m+2}\equiv16(-1)\pmod{60}\equiv-16+60$$
