Modular arithmetic problem: $7^x equiv 1 pmod{26}$

I have the following maths question, which I would like to solve in preparation for an exam:

"Find the smallest positive integer $x$ such that $7^x \equiv 1 \pmod{26}$. Calculate $7^{100} \bmod{26}$ (give your answer as a positive integer less than $26$)."

Any help would be much appreciated, even if it's just a little hint to the right approach to take, because I'm really stumped on this one.

Answer

As $26=2\cdot13$

we need $7^x\equiv1\pmod{13}$ and $7^x\equiv1\pmod2$

The second statement holds true for all integer $x$

So, the problem boils down to $7^x\equiv1\pmod{13}$

Now $x$ must divide $\phi(13)=12$

$7^2\equiv-3\pmod{13},7^3\equiv-21\equiv5,7^4\equiv(-3)^2\equiv-4,7^6\equiv5^2\equiv-1$

$\implies x\equiv0\pmod{12}$