Thats a good question. To answer this lets start with Clifford algebra generated by $\gamma$ matrices.
\begin{equation} \gamma_{\mu}\gamma_{\nu}+ \gamma_{\mu}\gamma_{\nu}=2\eta_{\mu\nu} \end{equation}
with $\mu,\nu=0,1,2,\cdots N$
with the metric signature $\eta_{\mu\nu}=\text{diag}(+,-,-,-,\cdots,-)$. Using $I$ and $\gamma_{\mu}$ we can construct a set of matrices as follow
\begin{equation} I, \gamma_{\mu},\gamma_{\mu}\gamma_{\nu}\quad(\mu<\nu), \gamma_{\mu}\gamma_{\nu}\gamma_{\lambda}\quad(\mu<\nu<\lambda),\cdots,\gamma_{1}\gamma_{2}\cdots\gamma_{N}. \end{equation}
There are \begin{equation} \sum_{p=0}^{N}\binom{N}{p} = 2^{N} \end{equation}
such matrices. Lets call them $\Gamma_{A}$, where $A$ runs from $0$ to $2^{N}-1$.
Now let $\gamma_{\mu}$ are $d\times d$ dimensional irreducible matrices. Our goal is to find a relation between $d$ and $N$. To this end lets define a matrix
\begin{equation} S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} \end{equation}. Where $Y$ is some arbitary $d\times d$ matrix. It is follows that
\begin{equation} (\Gamma_{B})^{-1}S\Gamma_{B} = \sum_{A=0}^{2^N-1}(\Gamma_{A}\Gamma_{B})^{-1}Y\Gamma_{A}\Gamma_{B} =\sum_{C=0}^{2^N-1}(\Gamma_{C})^{-1}Y\Gamma_{C}=S \end{equation}
Where we have used $\Gamma_{A}\Gamma_{B}=\epsilon_{AB}\Gamma_{C}$, with $\epsilon_{AB}^{2}=1$
Hence \begin{equation}S\Gamma_{A}=\Gamma_{A}S\end{equation}
Since $S$ commutes with all the matrices in the set, by Schur's lemma we conclude that $S$ must be proportional to the identity matrix so that we can write
\begin{equation} S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \lambda I \end{equation}
Taking trace we get
\begin{eqnarray} \text{Tr} S & = & \sum_{A=0}^{2^N-1} \text{Tr} Y = \lambda d\\
\Rightarrow \lambda & = & \frac{2^{N}}{d}\text{Tr} Y
\end{eqnarray}
or \begin{equation} \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \frac{2^{N}}{d}\text{Tr} Y \end{equation}
Taking the $(j; m)$ matrix element of both sides of last equation yield
\begin{equation} \sum_{A=0}^{2^N-1}((\Gamma_{A})^{-1})_{jk}(\Gamma_{A})_{km} = \frac{2^{N}}{d}\delta_{jm} \delta_{kl} \end{equation}
where $j; k; l; m = 1; 2;\cdots; d$ and we have used the fact that Y is an
arbitrary $d \times d$ matrix. If we set $j = k; l = m$ and sum over these
two indices, that gives
\begin{equation} \sum_{A=0}^{2^N-1} \text{Tr}[(\Gamma_{A})^{-1}] \text{Tr}[\Gamma_{A}] = 2^{N}\end{equation}
There are two cases to consider, namely, $N$ even and $N$ odd. For $N = 2M$ (even), $\text{Tr} \Gamma_{A} = 0$ except for $\Gamma_{0} = 1$ for which $\text{Tr} \Gamma_{0} = d$. Which gives
\begin{equation} d^2 = 2^N\qquad \text{or} \quad \boxed{d = 2^{N/2}} \end{equation}
This is the main result. For four dimensional Minkowski space time $N=4$ cosequntly the dimension of irreducible representation is $d = 2^{4/2} =4$.
