= Solution
This part proves $(a)\Rightarrow(b)$. If $M\ne0$ were a finite-dimensional unital <module>, its action would give a homomorphism
$$
A_n(k)\longrightarrow\operatorname{End}_k(M).
$$
The kernel is a two-sided <ideal> and is proper because the identity acts as the identity on $M$. A <simple ring> therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered <monomials> established in part (a), whereas the endomorphism <algebra> has dimension $(\dim_kM)^2$. Thus simplicity excludes every nonzero finite-dimensional <module>.
There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its <trace> gives $0=\dim_kM$ in $k$. In characteristic zero this forces $M=0$. This agrees with the <Weyl algebra has no nonzero finite-dimensional modules> criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.
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