[proofplan] We construct $F$ by successive approximation: define a sequence of polynomials $F_1, F_2, \ldots$ with $F_k \equiv L \pmod{(X_1, \ldots, X_n)^2}$ and $e_1(F_k) \equiv F_k(e_2(X_1), \ldots, e_2(X_n)) \pmod{\deg k+1}$, then pass to the limit $F = \lim F_k$. At each stage, the correction needed to achieve agreement to one higher degree is uniquely determined by solving a linear congruence modulo $\pi$, which has a unique solution because the relevant map is a bijection on the residue field. [/proofplan]

[step:Set up the successive approximation and base case] Define $F_1 = L(X_1, \ldots, X_n) = \sum_{i=1}^n a_i X_i$. We construct a sequence of power series $F_k \in \mathcal{O}_K[[X_1, \ldots, X_n]]$ for $k \geq 1$ satisfying: (i) $F_k \equiv L \pmod{(X_1, \ldots, X_n)^2}$, (ii) $F_k \equiv F_{k-1} \pmod{(X_1, \ldots, X_n)^k}$, (iii) $e_1(F_k) \equiv F_k(e_2(X_1), \ldots, e_2(X_n)) \pmod{(X_1, \ldots, X_n)^{k+1}}$. The base case $k = 1$: condition (iii) requires $e_1(F_1) \equiv F_1(e_2(X_1), \ldots, e_2(X_n)) \pmod{(X_1, \ldots, X_n)^2}$. Since $e_1, e_2 \in \mathcal{E}_\pi$, both satisfy $e(X) \equiv \pi X \pmod{X^2}$. The LHS is $e_1(L) \equiv \pi L \pmod{\deg 2}$, and the RHS is $L(e_2(X_1), \ldots, e_2(X_n)) \equiv L(\pi X_1, \ldots, \pi X_n) = \pi L \pmod{\deg 2}$. So condition (iii) holds for $k = 1$. [/step]

[step:Perform the inductive step: lift $F_k$ to $F_{k+1}$]Assume $F_k$ satisfies (i)--(iii). We seek $F_{k+1} = F_k + G_{k+1}$, where $G_{k+1}$ is a homogeneous polynomial of degree $k + 1$ in $X_1, \ldots, X_n$ with coefficients in $\mathcal{O}_K$, such that condition (iii) holds with $k$ replaced by $k + 1$. Define the error term \begin{align*} \Delta_k := e_1(F_k) - F_k(e_2(X_1), \ldots, e_2(X_n)). \end{align*} By hypothesis, $\Delta_k \equiv 0 \pmod{(X_1, \ldots, X_n)^{k+1}}$. Write $\Delta_k = D_{k+1} + O(\deg k+2)$, where $D_{k+1}$ is the homogeneous part of degree $k + 1$. We need $e_1(F_k + G_{k+1}) \equiv (F_k + G_{k+1})(e_2(X_1), \ldots, e_2(X_n)) \pmod{(X_1, \ldots, X_n)^{k+2}}$. Since $G_{k+1}$ is homogeneous of degree $k + 1$: - $e_1(F_k + G_{k+1}) \equiv e_1(F_k) + \pi G_{k+1} \pmod{\deg k+2}$, because $e_1(X) \equiv \pi X \pmod{X^2}$ and the chain rule contributes $\pi \cdot G_{k+1}$ at leading order. - $(F_k + G_{k+1})(e_2(X_1), \ldots, e_2(X_n)) \equiv F_k(e_2(X_1), \ldots, e_2(X_n)) + \pi^{k+1} G_{k+1} \pmod{\deg k+2}$, because $e_2(X_i) \equiv \pi X_i \pmod{X_i^2}$ and substituting into a homogeneous polynomial of degree $k + 1$ extracts the factor $\pi^{k+1}$. Therefore the new error at degree $k + 1$ is \begin{align*} D_{k+1} + \pi G_{k+1} - \pi^{k+1} G_{k+1} = D_{k+1} + (\pi - \pi^{k+1}) G_{k+1}. \end{align*} Setting this to zero gives \begin{align*} G_{k+1} = -\frac{D_{k+1}}{\pi - \pi^{k+1}} = -\frac{D_{k+1}}{\pi(1 - \pi^k)}. \end{align*}[/step]

[guided]The key point is that $1 - \pi^k$ is a unit in $\mathcal{O}_K$ for $k \geq 1$ (since $v_K(\pi^k) = k \geq 1 > 0$, so $v_K(1 - \pi^k) = 0$), and $\pi$ divides $D_{k+1}$. Why does $\pi$ divide $D_{k+1}$? Both $e_1$ and $e_2$ belong to $\mathcal{E}_\pi$, so $e_1(X) \equiv e_2(X) \equiv X^q \pmod{\pi}$. Working modulo $\pi$, the Frobenius endomorphism $X \mapsto X^q$ satisfies $\bar{e}_1(\bar{F}_k) = \bar{F}_k^q$ and $\bar{F}_k(\bar{e}_2(X_1), \ldots, \bar{e}_2(X_n)) = \bar{F}_k(X_1^q, \ldots, X_n^q) = \bar{F}_k^q$ (since raising to the $q$-th power is a ring homomorphism in characteristic $p$). So modulo $\pi$, the error $\Delta_k \equiv 0$, which means $\pi | D_{k+1}$. Since $\pi$ divides $D_{k+1}$ and $\pi(1 - \pi^k)$ is $\pi$ times a unit, the quotient $G_{k+1} = -D_{k+1} / (\pi(1 - \pi^k))$ has coefficients in $\mathcal{O}_K$, and the inductive step is complete.[/guided]

[step:Verify that $\pi$ divides $D_{k+1}$] We must confirm that $G_{k+1}$ has coefficients in $\mathcal{O}_K$. Since $1 - \pi^k \in \mathcal{O}_K^\times$ for $k \geq 1$ (as $v_K(\pi^k) = k \geq 1$ implies $v_K(1 - \pi^k) = 0$), it suffices to show $\pi \mid D_{k+1}$. Both $e_1, e_2 \in \mathcal{E}_\pi$ satisfy $e(X) \equiv X^q \pmod{\pi}$. Reducing modulo $\pi$ and working over $k_K = \mathcal{O}_K / \pi \mathcal{O}_K$: \begin{align*} \bar{e}_1(\bar{F}_k) = \bar{F}_k^q, \qquad \bar{F}_k(\bar{e}_2(X_1), \ldots, \bar{e}_2(X_n)) = \bar{F}_k(X_1^q, \ldots, X_n^q) = \bar{F}_k^q, \end{align*} where the last equality uses the Frobenius endomorphism: in characteristic $p$ with $q = p^f$, the map $x \mapsto x^q$ is a ring homomorphism, so $\bar{F}_k(X_1^q, \ldots, X_n^q) = \bar{F}_k(X_1, \ldots, X_n)^q$. Therefore $\bar{\Delta}_k = \bar{e}_1(\bar{F}_k) - \bar{F}_k^q = 0$ in $k_K[[X_1, \ldots, X_n]]$, which means $\pi \mid \Delta_k$ and in particular $\pi \mid D_{k+1}$. [/step]

[step:Pass to the limit and establish uniqueness] The sequence $(F_k)_{k \geq 1}$ satisfies $F_{k+1} \equiv F_k \pmod{(X_1, \ldots, X_n)^{k+1}}$, so it converges coefficient-wise to a power series \begin{align*} F = \lim_{k \to \infty} F_k \in \mathcal{O}_K[[X_1, \ldots, X_n]]. \end{align*} By construction, $F \equiv L \pmod{(X_1, \ldots, X_n)^2}$ and $e_1(F) = F(e_2(X_1), \ldots, e_2(X_n))$ (the identity holds at every degree). For uniqueness: if $F'$ is another solution, then $F - F' \equiv 0 \pmod{(X_1, \ldots, X_n)^2}$ and $e_1(F) - e_1(F') = (F - F')(e_2(X_1), \ldots, e_2(X_n))$. If $F - F' \not\equiv 0$, let $k + 1 \geq 2$ be the lowest degree where they differ, with homogeneous part $H_{k+1}$. The same degree-$(k+1)$ analysis as in the inductive step gives $\pi H_{k+1} = \pi^{k+1} H_{k+1}$, hence $(\pi - \pi^{k+1}) H_{k+1} = 0$. Since $\pi(1 - \pi^k)$ is nonzero in $\mathcal{O}_K$ (it has valuation $1$), we conclude $H_{k+1} = 0$, a contradiction. Therefore $F = F'$. [/step]