What is the nature of the mysterious Kw, the ionic product of water? In pure water, [H$^+$] = 10$^{-7}$ M and [OH$^-$] = 10$^{-7}$ M. So, we conclude pH of water is 7 and so is the pOH. We define the ionic product of water as Kw = [H$^+$][OH$^-$] = 10$^{-14}$ M$^{2}$. Till this, everything is fine.

Now, without any premise, it is concluded that the product [H$^+$][OH$^-$] of all concentrations in solutions of all sorts is the same, i.e. 10$^{-14}$. 

This leads to the conclusion that for all acidic, basic and neutral solutions, pH + pOH = 14.

So, why is it concluded that [H$^+$][OH$^-$] = 10$^{-14}$ M$^2$? What is the nature of Kw, the ionic product of water, that mysteriously is applicable to all solutions?
The solution product for the reaction $\ce{H2O -&gt; H+ + OH-}$ is a constant, $10^{-14}$, which we call $K_w = [\ce{H+}][\ce{OH-}]$. To make any aqueous solution, we put a solute into water. Because the dissociation constant (which is basically an equilibrium constant) of a substance only changes with temperature, $K_w$ remains the same. Thus, for any aqueous solution at a given temperature, $K_w$ is constant. 

You need to treat $K_w$ as an equilibrium constant. For example, if I have a reaction $\ce{A -&gt; 2B + C}$, $K_{eq} = \frac{[\ce{B}]^2[\ce{C}]}{[\ce{A}]}$; let's say I measured at equilibrium $[\ce{B}] = 1$ M, $[\ce{C}] = 2$ M, and $[\ce{A}] = 0.1$ M . This means $K_{eq} = 20$. Now, if I increased the concentration of $[\ce{A}]$ to $5$ M, $K_{eq}$ wouldn't change: the concentrations would have to adapt to my change. This is the same for water: once I determine $K_w$, it is the same for all concentrations of $\ce{H+}$ and $\ce{OH-}$.