Why does period 3 of the periodic table contain 8 elements instead of 18? Period 1 of the periodic table contains 2 elements ($1s^1$ and $1s^2$).

Period 2 contains 8 elements ($2s^1$, $2s^2$, $2p^1$, $2p^3$, ..., $2p^6$).

By the same argument, period 3 might contain 18 elements ($3s^1$, $3s^2$, $3p^1$, $3p^3$, ..., $3p^6$, $3d^1$, $3d^2$, $3d^{10}$). Why does period 3 of the periodic table contain 8 elements instead of 18?

I think that it could be explained by means of atomic stability, but I need help to understand what happens.

Thank you in advance.
You could say that the period number tells you about the largest value of principal quantum number in which a electron is present.

Electrons are filled according to $n+l$ rule. It states as $n+l$ increases, the energy of the orbital increases. If $n+l$ is the same, then the bigger $n$ has larger energy.

So $\mathrm{4s}$ would be filled before $\mathrm{3d}$. Therefore elements containing $\mathrm{3d}$ electrons would have their $\mathrm{4s}$ orbital filled, and therefore, come in 4th period. There 3rd period contains elements having $\mathrm{3s}$ and $\mathrm{3p}$ electron as their last electron which are equal to 8.