Show that $233$ divides $2^{29} - 1$ 
Show that $233$ divides $2^{29} - 1$.

At first I've tried to find some trivial congruence to brute force my way to the 29th  exponent, however it seems that only $2^{29} \equiv 1 \pmod{233} $. 
\begin{align} 2^8 &\equiv 23 &\pmod{233}\\
2^{12} &\equiv 135 &\pmod{233}\\
2^{16} &\equiv 63 &\pmod{233}\\
2^{24} &\equiv 51 &\pmod{233}\\
2^{29} &\equiv 1 &\pmod{233} \end{align}
Is there some clever congruence to make this brute force less... brute or is little Fermat a must in this situation?
 A: Exponentiation by squaring is the usual technique.  This isn't so different from what you did.  \begin{align*}
    2^1 &= 2 \pmod{233}  \\
    2^2 &= 4 \pmod{233}  \\
    2^4 &= 16 \pmod{233}  \\
    2^8 &= 23 \pmod{233}  \\
    2^{16} &= 23^2 \cong 23(10 + 10 + 3) \cong -3 -3 +69 \cong 63 \pmod{233}  \\
    \text{and } 29 &= 16+8+4+1 , \text{ so}  \\
    2^{29} &\cong 63 \cdot 23 \cdot 16 \cdot 2  \cong 46\,368 \cong 1 \pmod{233} \\
\end{align*}
Generally, there is no shortcut known for determining the order of an element modulo a prime.  (Here, we showed that the order of $2$ mod $233$ divides $29$.  Since $29$ is prime, and the order of $2$ is not $1$, its order is $29$.)
Addition-chain exponentiation is a method for shortcutting some of the above.  However, finding minimal addition chains is an NP-complete problem, so maybe this isn't the best choice for hand calculation.
A: Fermat's Little Theorem is pretty useful and probably the easiest way to solve this. Notice that $233$ is prime and that $29\mid 232$. Particularly, $\frac{232}{29}=8$ So, by Fermat's Little Theorem,
$$2^{232}\equiv 1\pmod{233}$$
and $232=29\cdot 8$ should get you to the solution.
A: As has already been pointed out, since $233$ is prime, we can by Fermat's Theorem infer that
$$ 2^{232} = 2^{(29)(8)} \equiv 1 \pmod{233}.$$
However, to get to the desired solution, observe that $ (2^{29})^{8} \equiv 1 \pmod{233}$. 
This means that either $2^{29} - 1 \equiv 0 \pmod{233} $ or $ 2^{29} + 1 \equiv 0 \pmod{233}$. 
Now, as $2^{29}$ is not a very large number, we can verify by calculator that $233 | 2^{29} - 1$.
