Answer:
Apply induction on [tex]n[/tex] (for integers [tex]n \ge 1[/tex]) after showing that [tex]\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)[/tex] is divisible by [tex]7[/tex] for [tex]j \in \lbrace 1,\, \dots,\, 6 \rbrace[/tex].
Step-by-step explanation:
Lemma: [tex]\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)[/tex] is divisible by [tex]7[/tex] for [tex]j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace[/tex].
Proof: assume that for some [tex]j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace[/tex], [tex]\genfrac{(}{)}{0}{}{7}{j}[/tex] is not divisible by [tex]7[/tex].
The combination [tex]\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)[/tex] is known to be an integer. Rewrite the factorial [tex]7![/tex] to obtain:
[tex]\displaystyle \begin{pmatrix}7 \\ j\end{pmatrix} = \frac{7!}{j! \, (7 - j)!} = \frac{7 \times 6!}{j!\, (7 - j)!}[/tex].
Note that [tex]7[/tex] (a prime number) is in the numerator of this expression for [tex]\genfrac{(}{)}{0}{}{7}{j}\![/tex]. Since all terms in this fraction are integers, the only way for [tex]\genfrac{(}{)}{0}{}{7}{j}[/tex] to be non-divisible by [tex]7\![/tex] is for the denominator [tex]j! \, (7 - j)![/tex] of this expression to be an integer multiple of [tex]7\!\![/tex].
However, since [tex]1 \le j \le 6[/tex], the prime number [tex]\!7[/tex] would not a factor of [tex]j![/tex]. Similarly, since [tex]1 \le 7 - j \le 6[/tex], the prime number [tex]7\![/tex] would not be a factor of [tex](7 - j)![/tex], either. Thus, [tex]j! \, (7 - j)![/tex] would not be an integer multiple of the prime number [tex]7[/tex]. Contradiction.
Proof of the original statement:
Base case: [tex]n = 1[/tex]. Indeed [tex]6 \times 1 + 1^{7} = 7[/tex] is divisible by [tex]7[/tex].
Induction step: assume that for some integer [tex]n \ge 1[/tex], [tex](6\, n + n^{7})[/tex] is divisible by [tex]7[/tex]. Need to show that [tex](6\, (n + 1) + (n + 1)^{7})[/tex] is also divisible by [tex]7\![/tex].
Fact (derived from the binomial theorem [tex](\ast)[/tex]):
[tex]\begin{aligned} & (n + 1)^{7} \\ &= \sum\limits_{j = 0}^{7} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] && (\ast)\\ &= \genfrac{(}{)}{0}{}{7}{0} \, n^{0} + \genfrac{(}{)}{0}{}{7}{7} \, n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] \\ &= 1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\end{aligned}[/tex].
Rewrite [tex](6\, (n + 1) + (n + 1)^{7})[/tex] using this fact:
[tex]\begin{aligned} & 6\, (n + 1) + (n + 1)^{7} \\ =\; & 6\, (n + 1) + \left(1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \\ =\; & 6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \end{aligned}[/tex].
For this particular [tex]n[/tex], [tex](6\, n + n^{7})[/tex] is divisible by [tex]7[/tex] by the induction hypothesis.
[tex]\sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right][/tex] is also divisible by [tex]7[/tex] since [tex]n[/tex] is an integer and (by lemma) each of the coefficients [tex]\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)[/tex] is divisible by [tex]7\![/tex].
Therefore, [tex]6\, (n + 1) + (n + 1)^{7}[/tex], which is equal to [tex]6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right)[/tex], is divisible by [tex]7[/tex].
In other words, for any integer [tex]n \ge 1[/tex], if [tex](6\, n + n^{7})[/tex] is divisible by [tex]7[/tex], then [tex]6\, (n + 1) + (n + 1)^{7}[/tex] would also be divisible by [tex]7\![/tex].
Therefore, [tex](6\, n + n^{7})[/tex] is divisible by [tex]7[/tex] for all integers [tex]n \ge 1[/tex].
