Proof for exponentiation in modular arithmetic If  $e$ is a natural number, then this is true:

$$m^e \bmod\ n = (m\bmod\ n)^e\bmod\ n$$

This is often used when encrypting, especially with RSA, since one can avoid directly calculating $m^e$, which can be a very big number.

However, I haven't been able to find any documentation/proof for this conjecture, can anyone give a source or explain it?
That is not a conjecture, that is basic number theory and follows from the fact that

$$(ab)\textrm{ mod } n = \big((a\textrm{ mod } n)\cdot(b\textrm{ mod } n)\big)\textrm{ mod } n$$

Write $a=k_an+r_a$ and $b=k_bn+r_b$ and thus $a \textrm{ mod }  n = r_a$ and $b \textrm{ mod }  n = r_b$. Plugging into the left hand side will give you:

\begin{align}\big((k_an+r_a)(k_bn+r_b)\big)\textrm{ mod }  n &amp;= \big((k_a k_b  n + k_a r_b+ k_br_a  )n + (r_a   r_b)\big)\textrm{ mod }  n \\ &amp;= (r_a  r_b)\textrm{ mod }  n\end{align}

The last step is because any multiple of $n$ clearly yields a zero remainder when divided by $n$.

Plug into the right hand side gives you $(r_a r_b ) \textrm{ mod }  n$. This shows the equality which is what you're looking for.

Viewing the exponentiation $m^e \textrm{ mod } n$ as $(m \cdot \ldots \cdot m) \mod n$ (multiplying $m$ with itself $e-1$ times), then you have what you are looking for.