Let $p$ be a prime number, let $d$ be an integer greater than $1$ and let $f\in\mathbb{Z}_p[X_1,\cdots,X_d]^d$.
I have already proved the following $p$-adic interpolation theorem:
Theorem 1. (Poonen) Let $I:=(X_1,\cdots,X_d)$ and let define the following positive integer: $$c:\left\{\begin{array}{cc}2&\textrm{, if }p=2\\1&\textrm{, otherwise}\end{array}\right..$$ If $f-I\in p^c\mathbb{Z}_p[X_1,\cdots,X_d]^d$, then for all $x\in{\mathbb{Z}_p}^d$, there exists $g\in\mathbb{Q}_p[[X]]^d$ such that: $$\forall n\in\mathbb{N},g(n)=f^n(x).$$
I would like to deduce from Theorem 1. and Strassmann's theorem the following version of the Skolem-Mahler-Lech theorem (which implies the general one by Cassel's theorem):
Theorem. (Skolem-Mahler-Lech) Let $(u_n)_{n\in\mathbb{N}}$ be a recurring sequence with coefficients in ${\mathbb{Z}_p}^\times$, then the following set: $$\{n\in\mathbb{N}\textrm{ s.t. }u_n=0\}$$ is the union of a finite set and a finite union of complete arithmetic progressions.
Proof. There exists $d\in\mathbb{N}_{\geqslant 1}$ and $(a_0,\cdots,a_{d-1})\in\left({\mathbb{Z}_p}^\times\right)^d$ such that: $$\forall n\in\mathbb{N},a_n=\sum_{i=0}^{d-1}a_iu_{n-i}.$$ Let $A$ be the matrix with $1$s on the upper-diagonal, the $a_i$s with increasing indexes on the last row and $0$s everywhere else and for all $n\in\mathbb{N}$ let define the following column matrix: $$v_n:=\begin{pmatrix}u_n\\\vdots\\u_{n+d-1}\end{pmatrix}.$$ Then, for all $n\in\mathbb{N}$, one has: $$v_n=A^nv_0.$$ The matrix got by reduction of the coefficients of $A$ mod $p\mathbb{Z}_p$ define an element of $\mathrm{GL}_d(\mathbb{Z}_p/p\mathbb{Z}_p)$. Therefore, there exists $N\in\mathbb{N}$ and a matrix $B$ such that: $$A^N=I_d+pB.$$ I don't know what to do from there. $\Box$
Any hint will be appreciated!
Some of you might be interested in the answer to this question, so I figure I would post it.
First, there exists $a_0,\cdots,a_{d-1}\in k$ with $a_0\neq 0$ such that: $$\forall n\in\mathbb{N},u_{n+d}=\sum_{i=0}^{d-1}a_iu_{n+i}.$$ Let $K$ be the extension generated by $u_0,\cdots,u_{d-1},a_0,\cdots,a_{d-1}$ over $\mathbb{Q}$, using Cassel's theorem, there exists a prime number $p\geqslant 3$ such that there exists an embedding $\alpha:K\hookrightarrow\mathbb{Q}_p$ which satisfies: $$\alpha(u_0),\cdots,\alpha(u_{d-1}),\alpha(a_0),\alpha(a_0)^{-1},\alpha(a_1),\cdots,\alpha(a_{d-1})\in\mathbb{Z}_p.$$ Therefore, one may assume that $K$ is a subfield of $\mathbb{Q}_p$ and that $(u_n)_{n\in\mathbb{N}}$ is a sequence of $\mathbb{Z}_p$. From there, let define the following elements: $$\forall n\in\mathbb{N},v_n:=\begin{pmatrix}u_n\\\vdots\\u_{n+d-1}\end{pmatrix}.$$ It is well know that one has: $$\forall n\in\mathbb{N},v_n=A^nv_0.$$ Let $f$ be the element of $\mathbb{Z}_p[X_1,\cdots,X_d]^d$ define by the left multiplication of $A$ by the column formed by the variables $X_1,\cdots,X_d$. This way, one gets: $$\forall n\in\mathbb{N},v_n=f^n(v_0).$$ Let $\overline{A}$ be the reduction mod $p\mathbb{Z}_p$ of $A$, one has that $\overline{A}\in\textrm{GL}_d(\mathbb{Z}_p/p\mathbb{Z}_p)$ since $a_0\in\mathbb{Z}_p\setminus p\mathbb{Z}_p$ and $p\mathbb{Z_p}$ is maximal in $\mathbb{Z}_p$. Besides, $\textrm{GL}_d(\mathbb{Z}_p/p\mathbb{Z}_p)$ is of finite order and there exists $m\in\mathbb{N}$ and a matrix $B$ with coefficients in $\mathbb{Z}_p$ such that: $$A^m=I_d+pB.$$ Multiplying this equality by the column formed by $X_1,\cdots, X_d$ gives $h\in\mathbb{Z}_p[X_1,\cdots,X_d]^d$ such that: $$f-I=ph.$$ Using Poonen's theorem, for all $i\in\{0,\cdots,m-1\}$, there exists $g_i\in\mathbb{Q}_p[[X]]^d$ which is convergent on $\mathbb{Z}_p$ and is such that: $$\forall n\in\mathbb{N},g(n)=f^{mn}(v_i)=v_{mn+i}.$$ Finally, according to Strassmann's theorem, for each $i\in\{0,\cdots,m-1\}$ and $j\in\{1,\cdots,d\}$, one has the following dichotomy:
The $j$th component of $g_i$ is null, then: $$\forall n\in\mathbb{N},u_{mn+i+j-1}=0.$$
The $j$th component of $g_i$ has a finite number of zeros in $\mathbb{Z}_p$, which implies that: $$\#\{n\in\mathbb{N}\textrm{ s.t. }u_{mn+i+j-1}=0\}<\infty.$$ Whence the result, since $\{mn+i+j-1;n\in\mathbb{N}\}_{\substack{i\in\{0,\cdots,m-1\}\\j\in\{1,\cdots,d\}}}$ covers $\mathbb{N}$.