One option is to start out with the matrix representation for two sets of conjugate Grassmann numbers (see previous thread), $\theta_i, \pi_i$ with $i=1,...,N$, such that
$\{\theta_i,\theta_j\}=0,\quad\{\pi_{i},\pi_{j}\} = 0, \quad \{\theta_i,\pi_j\} = \delta_{ij}$
Then a $2N$-dimensional Clifford algebra can be built by
$\gamma_{i}=\theta_{i}+\pi_{i}\\
\gamma_{N+i}=i(\theta_{i}-\pi_{i})$
Given the above anti-commutation relations it is straightforward to verify that $\{\gamma_{i},\gamma_{j}\}=2\delta_{ij}\mathbf{1}$. For a odd number of dimensions the last $\gamma$-matrix can be found by considering the product
$\gamma_{2N+1} = i^N\prod_{i=1}^{2N}\gamma_{i} = i^N\gamma_{1}\gamma_2...\gamma_{2N}$
To get a representation of the Dirac algebra $\{\gamma_{\mu},\gamma_\nu\}=2g_{\mu\nu}\mathbf{1}$ with signature (+,-,-,...,-) simply rotate all but one of the matrices in the representation such that $\gamma_i\to i\gamma_i$ (and relabel a bit).
This approach enables one to derive general representations of the gamma matrices from Grassmann numbers.
However, another option exists, namely to start out with a lower dimensional representation of the Clifford algebra (which can be computed by the method described above). A well-known case of a lower-dimensional representation, which was also known to Weyl & Dirac, would be the Pauli matrices:
$\sigma_{1} = \left[\begin{matrix}
0 & 1 \\
1 & 0
\end{matrix}\right], \quad
\sigma_{2} = \left[\begin{matrix}
0 & -i \\
i & 0
\end{matrix}\right], \quad
\sigma_{3} = \left[\begin{matrix}
1 & 0 \\
0 & -1
\end{matrix}\right] $
From these matrices outer products, $\rho_i = \mathbf{1}\otimes \sigma_i$ and $\eta_i = \sigma_i \otimes \mathbf{1}$, can be formed. It is then clear that $[\rho_i,\eta_j]=0$ which makes it possible to choose five matrices from the set $\{\rho_i,\eta_j,\rho_i\eta_j\}$ which fulfill the Clifford algebra.
To make this approach a bit more explicit, consider starting with a diagonal matrix from the initial set for simplicity - let us choose $\rho_3$ ($\eta_3$ would have been another option). This leaves us with two potential sets of matrices, namely $\{\rho_1,\rho_2\eta_1,\rho_2\eta_2,\rho_2\eta_3\}$ and $\{\rho_{2},\rho_{1}\eta_{1},\rho_{1}\eta_{2},\rho_{1}\eta_{3}\}$. Since $\rho_1$ is real I choose the first set, making the matrices:
\begin{align}
\gamma_0 &= \left[\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{matrix}\right] = \rho_3,
&&\gamma_1 = \left[\begin{matrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{matrix}\right] = i\rho_2\eta_1 \\
\gamma_2 &= \left[\begin{matrix}
0 & 0 & 0 & -i \\
0 & 0 & i & 0 \\
0 & i & 0 & 0 \\
-i & 0 & 0 & 0
\end{matrix}\right] = i\rho_2\eta_2,
&&\gamma_3 = \left[\begin{matrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{matrix}\right] = i\rho_2\eta_3\\
\gamma_5 &= \left[\begin{matrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{matrix}\right] = \rho_1
\end{align}
where I have taken the liberty to rotate three of them as described above. In this way the Dirac representation is found. Notice that a few choices were made along the way but that several of them can be motivated by the search for a simple representation (choosing diagonal and/or real when possible).
This approach can naturally be generalized to generate higher dimensional representations as well.
Best Answer
The defining relation for the Clifford algebra, $Cl(1,d)$ is $$ \{\gamma_\mu,\gamma_\nu\}=2 \eta_{\mu\nu}\ \mathbf{1}\ , $$ For simplicity, I will assume that $\eta_{\mu\nu}=\text{Diag}(1,-1,\ldots,-1)$ with $\mu,\nu=0,1,\ldots,d$. Other signatures can easily be incorporated. It is easy to see that $\gamma_0^2=-\gamma_i^2=\mathbf{1}$ for $i=1,\ldots,d$. Using the defining relation, one has $$ \gamma_0 \gamma_i + \gamma_i \gamma_0 =0 \ . $$ Multiply the above equation by $\gamma_0$ and then take the trace to obtain $$ \text{Tr}(\gamma_i) + \text{Tr}(\gamma_0 \gamma_i \gamma_0)=0\implies \text{Tr}(\gamma_i)=0\ , $$ on using the cyclic property of the trace. Similarly, one can show $\text{Tr}(\gamma_0)=0$. So the defining property proves the tracelessness of the Dirac matrices.
Two representations, $\gamma_\mu$ and $\gamma_\mu'$, of the Clifford algebra are said to be equivalent if $\gamma_\mu' = S \cdot \gamma_\mu S^{-1}$ for some invertible matrix $S$.
Appendix A of the Physics Reports article by Sohnius might be a good starting point for the other properties.