I} t(t-1)\cdots (t-i+1) \rho_i(0)$, is a polynomial in $t$ whose leading term is $\rho_{\max I}(0) t^{\max I}$. Since $\max I \in I$, we have $\rho_{\max I}(0) \neq 0$, and so for large enough $t$, this leading term dominates the polynomial: there exists $t_0$ such that for all $t \geq t_0$, $$\Delta(x^t) = (\textrm{nonzero})x^{t-s} + \textrm{higher order terms}.$$ Since $k$ is of characteristic zero, it follows that if $t \geq t_0$, given any $g \in R$ such that $\nu(g) \geq t-s$ (that is, $g \in \mathfrak{m}^{t-s}$, where $\mathfrak{m} \subset R$ is the maximal ideal), we can solve the equation $\Delta(f) = g$ uniquely for $f \in R$ by recursion on the coefficients, and the unique solution $f$ belongs to $\mathfrak{m}^t$. Therefore the restriction of $\Delta$ is an isomorphism of $k$-spaces $\mathfrak{m}^t \xrightarrow{\sim} \mathfrak{m}^{t-s}$ for any $t \geq t_0$. Fix such a $t$, and consider the commutative diagram $$\begin{CD}
0 @>>> \mathfrak{m}^t @>>> R @>>> R/\mathfrak{m}^t @>>> 0\\
@.            @VV \Delta V     @VV \Delta V   @VV \overline{\Delta} V    @.\\
0 @>>> \mathfrak{m}^{t-s} @>>> R @>>> R/\mathfrak{m}^{t-s} @>>> 0\\
\end{CD}$$ of $k$-spaces with exact rows, where $\overline{\Delta}$ is the map induced on quotients by $\Delta$. We know that the left vertical arrow is 