Sketch of the proof: Let $\Gamma \le {\rm Iso}(\Bbb R^2)$ be a wallpaper group. Then $\Gamma$ has a normal subgroup isomorphic
to $\Bbb Z^2$ with finite quotient $F$. This finite group acts on the lattice $\Bbb Z^2$ by conjugation.
We obtain a faithful representation
$$
F \hookrightarrow {\rm Aut}(\Bbb Z^2)\cong GL_2(\Bbb Z).
$$
The group $GL_2(\Bbb Z)$ has exactly $13$ different conjugacy classes of finite subgroups,
called arithmetic ornament classes:
\begin{align*}
C_1 & \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right\rangle,\;
C_2 \cong \left\langle \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle,\;
C_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} \right\rangle, \\
C_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\rangle, \;
C_6 \cong \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix} \right\rangle, \;
D_1 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right\rangle, \\
D_1 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\rangle, \;
D_2 \cong \left\langle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\right\rangle, \\
D_2 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\right\rangle, \; D_3 \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}\right\rangle, \\
D_3 & \cong \left\langle \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}
\right\rangle,\\
D_4 & \cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
\right\rangle, \; D_6\cong \left\langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}\right\rangle.
\end{align*}
This is an easy computation. Here $C_1,C_2,C_3,C_4,C_6$ are cyclic groups and $D_1,D_2,D_3,D_4,D_6$ are dihedral groups. We use here, that the order $n$ of a subgroup must satisfy $\phi(n)=deg(\Phi_n)\mid 2$, so that $n=1,2,3,4,6$. This is called the crystallographic condition. The wallpaper groups arise from these $13$ classes by equivalence classes of extensions
$$
1\rightarrow \Bbb Z^2\rightarrow \Gamma\rightarrow F\rightarrow 1,
$$
determined by $H^2(F,\Bbb Z^2)$.
By computing $H^2(F,\Bbb Z^2)$ in each case we obtain $18$ inequivalent extensions, because in $13$ cases the cohomology is trivial, and in three cases we get $C_2,C_2$ and $C_2\times C_2$, i.e., $5$ additional possibilities, so that $13+5=18$.
This yields $17$ different groups, because two of them turn out to be isomorphic.
