According to the given information there are two integers $x$ and $y$ s.t. $(1) \;\; M = 10x+y$, where $10^{6n-1} \leq x < 10^{6n}$ and $0 \leq y < 10$. Moving the last digit $y$ of $M$ to the beginning of $M$, we obtain the number $(2) \;\; N = y \cdot 10^{6n-1} + x$. Assume $7|M$. By Fermats little theorem $7 \mid 10^6 - 1$, which implies $7 \mid 10^{6n} - 1$. This combined with formulas (1) and (2) give us $N \equiv -10x \cdot 10^{6n-1} + x = -x(10^{6n} - 1) \equiv -x \cdot 0 = 0 \pmod{7}$, i.e. $7|N$. q.e.d.