Intuition for why effective nuclear charge increases down a column
Initially, let's take the electrons as probability density fields instead of classical particles.

The increase on the number of protons in the nucleus increases the electric force on all electrons, but the core electrons suffer higher attraction due to higher proximity and lighter electronic blindage, so their probability density field is forced nearer the nucleus. 

Reminding of Electric Force:
$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$

Consequently, the outer electrons will suffer lower blindage force from these core electrons, parallelly to the nucleus attraction, allowing them to get closer to the nucleus as well.

As effect, of course, the outer electrons also force the inner electrons towards the nucleus, but their probability densities reach a minimum repulsion energy in such a distance that electrons on the same level are forced closer to the nucleus than on the atomic element above, going down a column in the periodic table.