jmc

Let us look at the powers of $5$: $$\begin{array}{rl}{5}^1& \equiv 5(\mathrm{mod}7)\\ 5^2& \equiv 4(\mathrm{mod}7)\\ 5^3& \equiv 6(\mathrm{mod}7)\\ 5^4& \equiv 2(\mathrm{mod}7)\\ 5^5& \equiv 3(\mathrm{mod}7)\\ 5^6& \equiv 1(\mathrm{mod}7).\end{array}$$ Since $5^6\equiv 1(\mathrm{mod}7),$ we see that $5^{30}\equiv (5^6{)}^5\equiv 1(\mathrm{mod}7),$ hence our desired remainder is $\boxed{1}.$