Choose a presentation $$ A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m) $$ and $0 \leq c \leq \min(n, m)$ such that (\ref{equation-strictly-standard-one}) holds for $\pi$ and such that \begin{equation} \label{equation-star} \pi f_{c + j} \in (f_1, \ldots, f_c) + (f_1, \ldots, f_m)^2 \end{equation} for $j = 1, \ldots, m - c$. Say $\rho$ maps $x_i$ to the class of $r_i \in R$. Then we can replace $x_i$ by $x_i - r_i$. Hence we may assume $\rho(x_i) = 0$ in $R/\pi^4 R$. This implies that $f_j(0) \in \pi^4R$ and that $A \to \Lambda$ maps $x_i$ to $\pi^4\lambda_i$ for some $\lambda_i \in \Lambda$. Write $$ f_j = f_j(0) + \sum\nolimits_{i = 1, \ldots, n} r_{ji} x_i + \text{h.o.t.} $$ This implies that the constant term of $\partial f_j/\partial x_i$ is $r_{ji}$. Apply $\rho$ to (\ref{equation-strictly-standard-one}) for $\pi$ and we see that $$ \pi = \sum\nolimits_{I \subset \{1, \ldots, n\},\ |I| = c} r_I \det(r_{ji})_{j = 1, \ldots, c,\ i \in I} \bmod \pi^4R $$ for some $r_I \in R$. Thus we have $$ u\pi = \sum\nolimits_{I \subset \{1, \ldots, n\},\ |I| = c} r_I \det(r_{ji})_{j = 1, \ldots, c,\ i \in I} $$ for some $u \in 1 + \pi^3R$. By Algebra, Lemma \ref{algebra-lemma-matrix-left-inverse} this implies there exists a $n \times c$ matrix $(s_{ik})$ such that $$ u\pi \delta_{jk} = \sum\nolimits_{i = 1, \ldots, n} r_{ji}s_{ik}\quad \text{for all } j, k = 1, \ldots, c $$ (Kronecker delta). We introduce auxiliary variables $v_1, \ldots, v_c, w_1, \ldots, w_n$ and we set $$ h_i = x_i - \pi^2 \sum\nolimits_{j = 1, \ldots c} s_{ij} v_j - \pi^3 w_i $$ In the following we will use that $$ R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (h_1, \ldots, h_n) = R[v_1, \ldots, v_c, w_1, \ldots, w_n] $$ without further mention. In $R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (h_1, \ldots, h_n)$ we have \begin{align*} f_j & = f_j(x_1 - h_1, \ldots, x_n - h_n) \\ & = \pi^2 \sum\nolimits_{k = 1}^c \left(\sum\nolimits_{i = 1}^n r_{ji} s_{ik}\right) v_k + \pi^3 \sum\nolimits_{i = 1}^n r_{ji}w_i \bmod \pi^4 \\ & = \pi^3 v_j + \pi^3 \sum\nolimits_{i = 1}^n r_{ji}w_i \bmod \pi^4 \end{align*} for $1 \leq j \leq c$. Hence we can choose elements $g_j \in R[v_1, \ldots, v_c, w_1, \ldots, w_n]$ such that $g_j = v_j + \sum r_{ji}w_i \bmod \pi$ and such that $f_j = \pi^3 g_j$ in the $R$-algebra $R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (h_1, \ldots, h_n)$. We set $$ B = R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (f_1, \ldots, f_m, h_1, \ldots, h_n, g_1, \ldots, g_c). $$ The map $A \to B$ is clear. We define $B \to \Lambda$ by mapping $x_i \to \pi^4\lambda_i$, $v_i \mapsto 0$, and $w_i \mapsto \pi \lambda_i$. Then it is clear that the elements $f_j$ and $h_i$ are mapped to zero in $\Lambda$. Moreover, it is clear that $g_i$ is mapped to an element $t$ of $\pi\Lambda$ such that $\pi^3t = 0$ (as $f_i = \pi^3 g_i$ modulo the ideal generated by the $h$'s). Hence our assumption that $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$ implies that $t = 0$. Thus we are done if we can prove the statement about smoothness. \medskip\noindent Note that $B_\pi \cong A_\pi[v_1, \ldots, v_c]$ because the equations $g_i = 0$ are implied by $f_i = 0$. Hence $B_\pi$ is smooth over $R$ as $A_\pi$ is smooth over $R$ by the assumption that $\pi$ is strictly standard in $A$ over $R$, see Lemma \ref{lemma-elkik}. \medskip\noindent Set $B' = R[v_1, \ldots, v_c, w_1, \ldots, w_n]/(g_1, \ldots, g_c)$. As $g_i = v_i + \sum r_{ji}w_i \bmod \pi$ we see that $B'/\pi B' = R/\pi R[w_1, \ldots, w_n]$. Hence $R \to B'$ is smooth of relative dimension $n$ at every point of $V(\pi)$ by Algebra, Lemmas \ref{algebra-lemma-localize-relative-complete-intersection} and \ref{algebra-lemma-flat-fibre-smooth} (the first lemma shows it is syntomic at those primes, in particular flat, whereupon the second lemma shows it is smooth). \medskip\noindent Let $\mathfrak q \subset B$ be a prime with $\pi \in \mathfrak q$ and for some $r \in \mathfrak a$, $r \not \in \mathfrak q$. Denote $\mathfrak q' = B' \cap \mathfrak q$. We claim the surjection $B' \to B$ induces an isomorphism of local rings $(B')_{\mathfrak q'} \to B_\mathfrak q$. This will conclude the proof of the lemma. Note that $B_\mathfrak q$ is the quotient of $(B')_{\mathfrak q'}$ by the ideal generated by $f_{c + j}$, $j = 1, \ldots, m - c$. We observe two things: first the image of $f_{c + j}$ in $(B')_{\mathfrak q'}$ is divisible by $\pi^2$ and second the image of $\pi f_{c + j}$ in $(B')_{\mathfrak q'}$ can be written as $\sum b_{j_1 j_2} f_{c + j_1}f_{c + j_2}$ by (\ref{equation-star}). Thus we see that the image of each $\pi f_{c + j}$ is contained in the ideal generated by the elements $\pi^2 f_{c + j'}$. Hence $\pi f_{c + j} = 0$ in $(B')_{\mathfrak q'}$ as this is a Noetherian local ring, see Algebra, Lemma \ref{algebra-lemma-intersect-powers-ideal-module-zero}. As $R \to (B')_{\mathfrak q'}$ is flat we see that $$ \left(\text{Ann}_R(\pi^2)/\text{Ann}_R(\pi)\right) \otimes_R (B')_{\mathfrak q'} = \text{Ann}_{(B')_{\mathfrak q'}}(\pi^2)/\text{Ann}_{(B')_{\mathfrak q'}}(\pi) $$ Because $r \in \mathfrak a$ is invertible in $(B')_{\mathfrak q'}$ we see that this module is zero. Hence we see that the image of $f_{c + j}$ is zero in $(B')_{\mathfrak q'}$ as desired.