Show $k\mid 12$ with $2^k=1\bmod 13$ Let $k$ be the smallest positive integer such that $2^k=1\bmod 13$. Show that $k\mid 12$.
I'm not very good at proofs and I'm confused as how to prove this. I started by saying $2^k-1=13n$. But I don't know where to go from there.

$$
\begin{array}{c|c}
\hline
i & \text{ $2^i $ mod 13} \\
\hline
  1&2 \\
  2&4 \\
  3&8 \\
  4&3 \\
  5&6 \\
  6&12 \\
  7&11 \\
  8&9 \\
  9&5 \\
  10&10 \\
  11&7 \\
  12&1 \\
\hline
\end{array}
$$
First, for $0 \le k \le 12$, only $2^{12} \equiv 2^0 \equiv 1 \quad mod(13)$
Assume $k=12i+j$, where $0 \le j < 12$, then
$$2^k \equiv 2^{12i+j} \equiv 2^{12i} \cdot 2^j \equiv 1\cdot 2^j \equiv 1 \quad mod(13)$$
So $2^j=1$, $j=0$, and $k=12i, k|12$
(smallest positive integer of k is 12, obviously)
