regular local graded surjection has zero kernel

If $e=0$, the map is the identity of $k$. Suppose $e\ge 1$. If the homogeneous kernel were nonzero, it would contain a nonzero homogeneous polynomial $f$ of degree $a\ge 1$, because the degree-zero map is injective.

Put $P=k[X_1,\dots ,X_e]$. Since $P$ is a domain, multiplication by $f$ injects $P(-a)$ into $P$. Counting monomials of total degree at most $n$, for $n\ge a$, gives ${\sum }_{j=0}^n{\dim }_k(P\mathrm{/}(f){)}_j=\left(\genfrac{}{}{0px}{}{n+e}{e}\right)-\left(\genfrac{}{}{0px}{}{n-a+e}{e}\right)$. This polynomial has degree $e-1$: the terms of degree $e$ cancel.

The surjection $P\mathrm{/}(f)\to {\mathrm{gr}}_{\mathfrak{m}}R$ bounds ${\mathrm{\ell }}_R(R\mathrm{/}{\mathfrak{m}}^{n+1})$ by that count, because its filtration factors are precisely the graded pieces. But its eventual Hilbert–Samuel polynomial has degree $\dim R=e$ and positive leading coefficient (it is eventually positive). A polynomial of degree $e$ cannot be bounded by one of degree $e-1$ for all large $n$. Thus no such $f$ exists.