Show that $\forall n\in\mathbb{N}$, $\frac{(7n)!}{7^nn!}\equiv(-1)^n\pmod7$

Question

I want to prove that for all $n\in\mathbb{N}$, $\frac{(7n)!}{7^nn!}\equiv(-1)^n\pmod7$. The obvious answer seems to be induction. So I show easily that this holds for $n=0$, but then during the induction I get that $$\frac{(7n+7)!}{7^{n+1}(n+1)!}=\frac{(7n)!}{7^nn!}\cdot \frac{\prod_\limits{i=7n+1}^{7n+7}i}{7(n+1)}\equiv(-1)^n\cdot\frac{\prod_\limits{i=7n+1}^{7n+7}i}{7(n+1)}\pmod7$$ and I don't really know how to go from there.

Answer 1

$$\frac{\prod_\limits{i=7n+1}^{7n+7}i}{7(n+1)}=\prod_\limits{i=7n+1}^{7n+6}i.$$ Now $$\prod_\limits{i=7n+1}^{7n+6}i\equiv 1\times2\times\cdots\times5\times6\pmod 7\equiv-1\pmod 7.$$