= Solution
A one-dimensional commutative <formal group law> over the <p-adic integers> is a power series $\mathcal F(X,Y)\in\mathbb Z_p[[X,Y]]$ with
$$
\mathcal F(X,0)=X,\quad \mathcal F(0,Y)=Y,\quad
\mathcal F(X,Y)=\mathcal F(Y,X),
$$
$$
\mathcal F(\mathcal F(X,Y),Z)=\mathcal F(X,\mathcal F(Y,Z)).
$$
Its linear part is $X+Y$, and it has a unique <formal inverse> $i(T)=-T+O(T^2)$ satisfying $\mathcal F(T,i(T))=0$. A formal-group homomorphism is a series $h(T)\in T\mathbb Z_p[[T]]$ satisfying $h(\mathcal F(X,Y))=\mathcal F(h(X),h(Y))$.
Define $[n](T)$ by adding $T$ to itself $n$ times with the <formal group law>. Induction on the linear part gives
$$
[n](T)=nT+O(T^2).
$$
Associativity and commutativity make $[n]$ a formal-group homomorphism. Since $p\nmid n$, its linear coefficient is a unit in $\mathbb Z_p$. We now construct its <compositional inverse of a formal power series>, rather than merely assert that the nonzero coefficient is sufficient.
Write $[n](T)=nT+a_2T^2+\cdots$ and seek $g(T)=b_1T+b_2T^2+\cdots$ with $[n](g(T))=T$. The degree-one coefficient gives $b_1=n^{-1}$. At degree $k\ge2$, the coefficient is $nb_k$ plus a polynomial in the already chosen $b_1,\ldots,b_{k-1}$ and $a_2,\ldots,a_k$. Division by the unit $n$ uniquely chooses $b_k\in\mathbb Z_p$. This recursively constructs $g$ over the same ring. A right inverse is constructed in the same way; associativity of composition identifies it with $g$, so $g\circ[n]=[n]\circ g=T$.
To prove that $g$ respects the group law, apply $[n]$ to the two candidate series:
$$
[n]\bigl(\mathcal F(g(X),g(Y))\bigr)
=\mathcal F(X,Y)
=[n]\bigl(g(\mathcal F(X,Y))\bigr).
$$
Cancel by composing with $g$. The result is $\mathcal F(g(X),g(Y))=g(\mathcal F(X,Y))$. Thus
$$
\boxed{[n]:\mathcal F\longrightarrow\mathcal F\text{ is an isomorphism over }\mathbb Z_p.}
$$
This is the <multiplication isomorphism of a formal group law>, a special case of the <invertible morphism criterion for formal group laws>. If $p\mid n$, its linear coefficient is not a unit, so the same isomorphism assertion would fail.
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