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:

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*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?

A more general way would be to prove that $a^n-b^n=(a-b)(a^{n-1}+ba^{a-2}+\cdots+b^{n-1}a+ b^{n-1})$.
Another way would be to prove the binomial theorem $(a+b)^n=\sum_{n\ge i\ge 0}\binom{n}{i}a^ib^{n-i}$ and then expand $11^n=(7+4)^n$.
A third and alternate (unnecessary) way is to notice that $11^n-4^n$ is the number of ways of giving one of $11$ candies $\lbrace c_1,\cdots,c_{11}\rbrace$ to $n$ children such that not all of them get one of $c_1,c_2,c_3$ or $c_4$. No matter the distribution of candies, there is at least one kid who has only $7$ options for candy. So, $11^n-4^n$ is always divisible by $7$.
