I know that the hyperoctahedral group $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ has the presentation
$$\langle s_{\text{1}},\ldots,s_n\mid s_{\text{1}}^{\text{2}}=s_i^2=1, (s_1s_2)^4=(s_is_{i+1})^3=(s_is_j)^2=1~| ~2\leq i\leq n, i+2\leq j\leq n \rangle,$$
where the generator $s_1$ is the diagonal matrix with a $-1$ in the top left entry and $1$s on the rest of the diagonal entry and $s_i$ are the permutation matrix formed from the identity switching both the $i-1$st and $i$th rows and corresponding columns.
Where can I get a proof for this?
This is not an answer per se but a few places to look, according to "Generators and Relations for Discrete Groups (Reprint of Fourth Edition)," by H. S. M. Coxeter and W. O. J. Moser, page 90, are:
Nielsen, J. (1924b), "Die Isomorphismengruppen der freien Gruppen," Math. Ann. 91, 169-209. This has a result that the group "may be regarded as the symmetry group of the $n$-dimensional Cartesian frame."
Young, A. (1930), "On quantitative substitutional analysis, " Proc. London Math. Soc. (2), 31, 273-388. This is where the group got its name.
The next two together show how the group "can be generated by two generators":
Neumann, B. H. (1932), "Die Automorphismengruppe der freien Gruppen," Math. Ann. 107, 370-373.
Coxeter, H. S. M. and Todd, J. A. (1936), "Abstract definitions for the symmetric groups of the regular polytopes in terms of two generators. Part I: The complete groups," Proc. Cambridge Philos. Soc. 32, 199.
I hope this helps.