Intuitive explanation

Preliminary: A vector has a many components as elements of the vector space basis.

A Clifford algebra basis is generated by all (independent) products of the generators (in Dirac's equation case these are the $\gamma$'s).

There are as many $\gamma$'s as the dimension of the spacetime.

Intuitive explanation

Preliminary: A vector has a many components as elements of the vector space basis.

A Clifford algebra basis is generated by all (independent) products of the generators (in Dirac's equation case these are the $\gamma$'s).

The counting

There are as many $\gamma$'s as the dimension of the spacetime, and according to the definition the algebra includes a unit, $$\bigl\{\gamma^a,\gamma^b\bigr\} = 2 \eta^{ab}\mathbf{1}.$$

For any extra element the new basis consists of the previous basis elements plus the product of each of those by the extra element. This is the new basis has twice the elements. Therefore, $$\dim(\mathcal{C}\ell(n)) = 2^{n}.$$

In order to represent this algebra one needs "matrices" of $2^{n/2}\times 2^{n/2}$, which is not bad for even dimensional spacetimes.

Said that, the problem (which I don't intend to demonstrate) comes with odd dimensional spacetimes... however, intuitively again, this algebra can be represented by two copies of the co-dimension one algebra, i.e. one dimension less. This reason is why the minimal dimensionality for the representation of the $\gamma$'s is $$\dim(\gamma) = 2^{\lfloor n/2\rfloor}\times 2^{\lfloor n/2 \rfloor}.$$

If you wonder whether one can find a bigger representation of the $\gamma$'s, the answer is YES, but you will will end up with either a non-fundamental or a trivial extension.