The effects of Pascal's principle, discovered by the French physicist in his famous barrel experiment.
In the experiment, also named "Pascal's Paradox," Pascal inserted a $10$ m-long vertical pipe into a barrel filled with water. At that point, Pascal began pouring water into the vertical pipe until it filled the same pipe and observed an increase in pressure, which reached such an intensity that it broke the cask.
Mathematically, the principle can be described by the following formula: $$\Delta p =\rho g (\Delta h)\label{1} \tag{1}$$
Where $\Delta p$ is the change in hydrostatic pressure, measured in pascals, introduced into the cask and due to the weight of the fluid poured into the pipe; $ρ$ is the density of the fluid, measured in kilograms per cubic meter; $g$ is the acceleration of gravity; and $\Delta h$ is the height, measured in meters, reached by the fluid inside the pipe.
The formula is derived from Stevin's law applied to the barrel-pipe system. In this case, the pressure induced into the cask is given by $$ p_0 = \rho g h_0 $$ where $h_0$ is the height of the fluid in the pipe.
The pressure at any point in the fluid contained in the cask will then be given by $$p_f = \rho g \Delta h + p_0 = \rho g \Delta h + \rho g h_0 = \rho g (\Delta h + h_0)$$ where $p_f$ is the final pressure at any point (after the addition of water), $\Delta h $ is the change in height of the liquid, $p_0$ is the initial pressure at that point. Taking the pressure $p_0$ to the other side of the equation gives the initial formula \eqref{1}.
Is there an alternative simplified method to mathematically prove Pascal's barrel paradox?
The easiest way I can image, though not elementary, is to transform your equation \eqref{1} into an ordinary differential equation and solve it with initial condition given by Stevin's law $$\DeclareMathOperator{\Dm}{d\!} \Dm p = \rho \mathrm{g}\Dm h \implies \begin{cases} \dfrac{\Dm p}{\Dm t}=\rho \mathrm{g} \dfrac{\Dm h}{\Dm t}\\ \\ \left.p\right|_{t=0} =p_0 \triangleq \rho g h_0 \end{cases} $$ Then we have $$ \begin{align} \int\limits_{0}^{t}{\Dm p} &=\rho \mathrm{g} \int\limits_{0}^{t}{\Dm h}\\ &\Updownarrow\\ p(t)-p_0 &=\rho \mathrm{g} \big(h(t)-h_0\big)\\ &\Updownarrow\\ \text{and by putt}&\text{ing $\Delta h(t)=h(t)-h_0$}\\ &\Updownarrow\\ p(t)-p_0 &=\rho \mathrm{g} \Delta h(t)\\ &\Updownarrow\\ p(t)=\rho \mathrm{g} \Delta h(t) & +p_0 =\rho \mathrm{g} \big( \Delta h(t) +h_0\big) \end{align} $$ This way of doing things has also the advantage of showing that the change of pressure (approximatively since the model does not takes in to account the real equations of fluid motions) does not depend on the way we pour the water in the pipe, i.e. $h(t)$ can be any continuous function. Well, my two cents.