Degree $2$ nilpotent matrices with non-zero product Let $n$ be sufficiently large positive integer. 
Let $M_i$ be $n$ matrices with entries in either $\mathbb{C}$
or $\mathbb{Z}/n \mathbb{Z}$.

Is it possible, (1) for $1 \le i \le n, M_i^2=0$ and
  (2) every product of the permutations of $M_i$ doesn't vanish: $\prod_{i} M_i \ne 0$?

If it is possible, $M_i$ better be as small as possible.
Explicit construction for single $n$ as large as possible is of
interest to me.

For $n=3$ solution exists with 3 by 3 matrices,
coming from parametrization of nilpotent matrices.
In flattened form:
 M1=[-59/20, -767/400, -59/400, 3, 39/20, 3/20, 20, 13, 1]
 M2=[-441, -8379, -93933, 12, 228, 2556, 1, 19, 213]
 M3=[-1931, -17379, -44413, 212, 1908, 4876, 1, 9, 23]

 A: A general nilpotent matrix in dimension 2 can be written in the form
$$M(v):=\begin{pmatrix} xy & -x^2 \\ y^2 & -xy\end{pmatrix}, \quad v=\begin{pmatrix}x\\y\end{pmatrix}.$$ The best is to interpret $\mathfrak{sl}_2$ as $\mathfrak{sp}(V,\omega)$ where $(V,\omega)$ is a $2$-dimensional symplectic space. Then one has a natural isomorphism $\phi:V\to V^*$, and the matrix above, with suitable choice of the form, represents the map which corresponds to $v \otimes \phi(v)$ for some $v\in V$ (viewing endomorphisms as $V\otimes V^*$). Take matrices of this form with vectors $v_1$, $\ldots$, $v_n$ such that $\omega(v_i, v_j) \neq 0$ for $i\neq j$ (if you want to do that for all $n$, this requires the field to be infinite).
A: Let $A^{ij}$, where $1\leq i,j\leq n$, be $n\times n$ matrices with elements given by
$$
(A^{ij})_{ab} = \delta_{ia}(1-\delta_{jb}),
$$
where $\delta$ is Kronecker delta. In other words, the matrix $A^{ij}$ is filled with zeros, except for its $i$-th row, which is filled with ones, except for its $j$-th place, which is again zero. Then
$$
A^{ij}A^{kl} = \begin{cases} 0 &\text{if} ~~j=k;\\ A^{il}& \text{if} ~~j\neq k.\end{cases}
$$
Set $M_i = A^{ii}$. Then $M_i^2 = 0$ and $\prod\limits_{k=1}^n M_{i_k} = A^{i_1i_n}$ if $i_1\neq i_2\neq \dots \neq i_n$.
