$1-\phi$ is surjective for a $\phi$-module?

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Let $A$ be a ring which is complete for the $p$-adic topology, in another word, there is an isomorphism between the two topological rings $A\rightarrow \lim_{\longleftarrow_{n \in \mathbb N_+}} A/P^nA$. Assume $A/pA=k$ with $k$ is an algebraic closed field.(it is easy to see the characteristic of $k$ is $p$), so $A$ is endowed with a frobenius $\sigma$ lifting the map $x\rightarrow x^p$ in $k$.

Define $V$ is a free $A$ module of rank $n$ and with a map $\phi: V\rightarrow V$ which satisifies $\phi(k_1v_1+k_2v_2)=\sigma(k_1)\phi(v_1)+\sigma(k_2)\phi(v_2)$ and there exists a basis such that $Mat(\phi)\in GL_n(A)$.

Then the question is how to prove $1-\phi$ is surjective. This is an important fact metioned in Laurent Berger's note about $p$-adic Galois representations(page 19, after the proof of corollary 8.7).

Thanks!