Archimedes principle for arbitrarily shaped bodies can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area to an integral over the enclosed volume. $$\oint p(\vec{r})\ d\vec{A} = \int \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.
Now, as the position-dependent function we take the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$).
Then the gradient of (2) is $$\nabla p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction.
Inserting (3) into (1) we get $$\oint p(\vec{r}) d\vec{A} = \int (-\rho g\hat{z})\ dV. $$
Now on the left side $p\ d\vec{A}$ obviously is (except for its sign) the pressure force acting on the surface area element $d\vec{A}$. And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint d\vec{F}=-\rho g \hat{z} \int dV$$ or finally $$\vec{F}=\rho g \hat{z} V$$ which is just Archimedes' principle.