= Solution
The proof of the <linearity of a uniform pro-p group> has three distinct steps: pass to a finite-dimensional <Lie algebra>, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful <Lie algebra representation> need not send the original integral lattice into the convergence domain of the <matrix exponential>.
Let $G$ be a <uniform pro-p group> and let $L_G$ be its intrinsic <powerful Lie lattice>. Then
$$
\mathfrak g=\mathbb Q_p\otimes_{\mathbb Z_p}L_G
$$
is a finite-dimensional <Lie algebra> over a <field> of <characteristic zero>. The <Ado theorem>, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional <Lie algebra representation>
$$
\rho:\mathfrak g\hookrightarrow\mathfrak{gl}(V),\qquad \dim_{\mathbb Q_p}V=m<\infty.
$$
One must use a faithful representation rather than just the <Adjoint representation>, which annihilates the <center of a Lie algebra>.
Choose a <p-adic lattice> $\Lambda\subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing <matrices> of a module basis have bounded denominators. Hence for sufficiently large $k$,
$$
\rho(p^kL_G)\subseteq p\operatorname{End}_{\mathbb Z_p}(\Lambda).
$$
For odd $p$, the <matrix exponential> and <p-adic logarithm> converge and are inverse between $p\operatorname{End}(\Lambda)$ and $1+p\operatorname{End}(\Lambda)$. Put $H=G^{p^k}$; under the <uniform pro-p group and powerful Lie lattice correspondence> it is the BCH group on $p^kL_G$. Define
$$
\tau:H\longrightarrow\operatorname{GL}(\Lambda),\qquad
\tau(\exp_G X)=\exp(\rho(X)).
$$
Here $\exp_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The <Lie algebra homomorphism> $\rho$ respects the convergent <Baker--Campbell--Hausdorff formula>, so
$$
\exp(\rho(X*Y))=\exp(\rho(X))\exp(\rho(Y)).
$$
Thus $\tau$ is a continuous <group homomorphism>. If $\tau(\exp_G X)=1$, taking the <matrix logarithm> gives $\rho(X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the <open normal subgroup> $H$.
To finish, use the <extension of a faithful representation from an open normal subgroup>. Choose representatives $t_1=1,t_2,\ldots,t_r$ of the finite left <cosets> $G/H$, where $r=[G:H]$. On the finite free module $W=\bigoplus_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquely
$$
gt_j=t_i h_{ij},\qquad h_{ij}\in H,
$$
and mapping the $j$-th block to the $i$-th block by $\tau(h_{ij})$. If $g't_j=t_\ell h'$ and $gt_\ell=t_i h$, then $gg't_j=t_i hh'$; this proves the representation identity because $\tau(hh')=\tau(h)\tau(h')$. Each block matrix is invertible over $\mathbb Z_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.
If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau(g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this <induced representation> is faithful. Choosing a basis of $W$ proves
$$
\boxed{G\hookrightarrow\operatorname{GL}_{mr}(\mathbb Z_p)}.
$$
The continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\operatorname{End}(\Lambda)$ instead; the preceding questions concern odd $p$.
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