Prove that $7 \mid 11^n - 4^n$ with mathematical induction I want to prove that $7 \mid 11^n - 4^n$ with mathematical induction. This is what I wrote:

*

*For $n = 1$, we have $7 \mid 11^1 - 4^1 \Rightarrow 7 \mid 7$ which is obviously true. $\checkmark$


*Assume that the statement is true for $n = k$. So: $7 \mid 11^k - 4^k$ and therefore there is an integer such that $m$ such that: $11^k - 4^k = 7m$. Thus: $4^k = 11^k - 7m$.


*Now we prove its truth for $n = k + 1$. So we want to prove that $7 \mid 11^{k + 1} - 4^{k + 1}$. We can write this like this: $7 \mid \big(11\cdot11^k\big) - \big(4\cdot4^k\big)$ and since $4^k = 11^k - 7m$, so in fact we have to prove: $7 \mid \big(11 \cdot11^k\big) - \big(4\cdot(11^k - 7m)\big) \Longrightarrow 7 \mid 11\cdot11^k - 4\cdot11^k + 7(4m) \Longrightarrow 7 \mid 7\cdot11^k + 7(4m) \Longrightarrow 7 \mid 7(11^k + 4m)$.
But from here you can clearly see that $7(11^k + 4m)$ is divisible by 7. So according to the principle of mathematical induction, the statement is proved. $\blacksquare$
Now my question is, is this proof correct? And is there another way to prove this statement by mathematical induction?

Let's modify your induction proof so that we directly use the induction hypothesis that $7 \mid 11^k - 4^k$ for some positive integer $k$.
Let $P(n)$ be the the statement that $7 \mid 11^n - 4^n$.
You handled the $n = 1$ case correctly.
Since the statement $7 \mid 11^n - 4^n$ holds for $n = 1$, we may assume $P(k)$ holds for some positive integer $k$.  That is, $7 \mid 11^k - 4^k$ holds for some positive integer $k$.  Then there exists an integer $m$ such that $7m = 11^k - 4^k$.
Thus far, I have essentially reformulated what you said in your proof.  Now, we will express $11^{k + 1} - 4^{k + 1}$ in terms of $11^k - 4^k$ so that we can use the induction hypothesis in the induction step.
Let $n = k + 1$.  Then
\begin{align*}
11^{k + 1} - 4^{k + 1} & = 11 \cdot 11^k - 4 \cdot 4^{k}\\
                       & = (7 + 4)11^k - 4 \cdot 4^k\\
                       & = 7 \cdot 11^k + 4 \cdot 11^k - 4 \cdot 4^k\\
                       & = 7 \cdot 11^k + 4(11^k - 4^k)\\
                       & = 7 \cdot 11^k + 4 \cdot 7m && \text{by the induction hypothesis}\\
                       & = 7(11^k + 4m)
\end{align*}
Since $11^k + 4m$ is an integer, $7 \mid 11^{k + 1} - 4^{k + 1}$.  Thus, $P(k) \implies P(k + 1)$ for each positive integer $k$.  Since $P(1)$ holds and $P(k) \implies P(k + 1)$ for each positive integer $k$, $P(n)$ holds for each positive integer $n$ by the Principle of Mathematical Induction.
