$376$ 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.
Proof is easiest by induction. You have stated your answer but you just need to word it right.
First off $$ 376^2 \equiv 376 \mod 1000 $$ can be verified directly.
The general statement as you point out is $$ 376^k\equiv 376 \mod 1000 $$
Assume it to be true for $k$. Now $$ 376^{k+1}\equiv 376^k \cdot 376 \equiv 376 \cdot 376 \equiv 376^2 \equiv 376\mod 1000 $$
This proves your statement.
Note: I am essentially rewriting what you state. I hope this is what you are looking for!
Hint $\,\ a^2\equiv a,\,\ \color{#c00}{a^n\equiv a}\ \Rightarrow\, a^{n+1}\equiv a\color{#c00}{a^n}\equiv a\color{#c00}a\equiv a$
