$-\Delta u =f $ for $f\in L^p(\Omega)$ implies $\nabla u \in L^p(\Omega)$

Question

I wonder if there is a quick&dirty proof of the following statement: Assume $\Omega \subset \mathbb{R}^n$ open and bounded with $C^1$-boundary and $p\geq 2$. The unique solution $u\in H^1_0(\Omega)$ of

$$-\Delta u=f ,\quad f\in L^p(\Omega)$$

is actually in $W^{1,p}(\Omega)$ and its $W^{1,p}$-norm can be bounded by $|f|_{L^p}$.