Exponentials

$13 \equiv 1 (\mod 12) \Rightarrow 13^{13} \equiv 1 (\mod 12)$ And $1 \equiv 1 (\mod 12)$ Therefore we have: $13^{13} -1 \equiv 1-1 (\mod 12) \Rightarrow 13^{13}-1 (\mod 12) = 0$

$13^{13} = (12+1)^{13} = \sum_{n=0}^{13}\binom{13}{n}12^{13-n}(1^n) = 1+\sum_{n=0}^{12}\binom{13}{n}12^{13-n}(1^n)$ Hence $13^{13}-1$ will have 12 as a factor, so $13^{13} \equiv 1\pmod{12}$

Direct result of FERMAT THEOREM