Hamiltonian of a particle in magnetic field squared

I’m trying to follow Tong lectures about Gauge Theories, but I think I’m doing some really stupid mistake.

At one point he takes the Hamiltonian for a spin $1/2$ particle in a potential as the usual
begin{equation*}
H=(mathbf{p}-emathbf{A})cdotmathbf{sigma}
end{equation*}
Where $mathbf{p}$ is the momentum, $mathbf{A}$ the vector potential and $mathbf{sigma}$ is the vector of the Pauli matrices.
Now he squares the result, using $sigma_isigma_j=delta_{ij}+iepsilon_{ijk}sigma_k$ and he finds
begin{equation*}
H^2=(mathbf{p}-emathbf{A})^2-2emathbf{B}cdotmathbf{S}
end{equation*}
with $mathbf{S}=frac{1}{2}mathbf{sigma}$ the spin operator.

I tried to reproduce this calculation, but I think I’m missing something. I write $H^2$ as
begin{align*}
H^2&=(p_i-eA_i)sigma_i(p_j-eA_j)sigma_j=
&=(p_i-eA_i)(p_j-eA_j)(delta_{ij}+iepsilon_{ijk}sigma_k)=
&=(mathbf{p}-emathbf{A})^2+iepsilon_{ijk}sigma_k(p_i-eA_i)(p_j-eA_j)
end{align*}
The last term seems to be the product of an antisymmetric tensor $epsilon_{ijk}$ in $ij$ times a symmetric part, and this should go to zero.

What am I missing? How do I recover the $-2emathbf{B}cdotmathbf{S}$ term?