Trying to remove a mod from an equation. $376$ is a number that for positive integers $n$, $376^n$ will always end with the number $376$. Now knowing that $376^k \mod 1000 = 376$.
How do you prove that the following is true.
$$
376^k \mod 1000 = 376^{k+1} \mod 1000
$$
Is there a way to cancel the mods or can you plug $376^k \mod 100 = 376$ into the equation somehow?

Of course it is true that
$$
a \mod b = a - \left\lfloor \frac{a}{b}\right\rfloor b
$$
but removing mod from your equation will make your life harder, not easier.  You need to do some number theory.
Consider the difference $376^{k+1} - 376^k$, for $k \ge 1$.
This is equal to $376^{k - 1} ( 376^2 - 376 ) $.
Now,
\begin{align*}
376^2 - 376
&= 376(376 - 1) \\
&= 376(375) \\
&= (8 \cdot 47)(125 \cdot 3) \\
&= (8 \cdot 125)(3 \cdot 47) = (1000) (3 \cdot 47)
\end{align*}
which is a multiple of $1000$.  So the difference between the two numbers is a multiple of $1000$;
hence they have the same remainder, which proves equation 1.
