Number of form 4666...6669 is Divisible by 7

Theorem
Let $x$ be a natural number in the form:
 * $\sqbrk {4 \underbrace {666 \cdots 6}_n 9}_{10}$

Then $x$ is divisible by $7$.

Proof
We have:

The sum of the digits in $14 \times 10^{n + 1} + 7$ is in fact $12 = 1 + 4 + 7$ which is a multiple of $3$.

Hence, by Divisibility by $3$, $\dfrac {14 \times 10^{n + 1} + 7} 3$ is indeed an integer.

Then: