This follows by a straightforward induction on \(\left\vert I\right\vert \).

Base case (\(\left\vert I\right\vert =0\)): The empty product is \(\underline{1}\) (the constant FPS with value \(1\)), and we have \(\underline{1}\circ g=\underline{1}\) for any \(g\in K\left[ \left[ x\right] \right] \). Hence both sides equal \(\underline{1}\).

Induction step: Assume the result holds for all sets of size \(\left\vert I\right\vert - 1\). Pick some \(a\in I\) and write \(\prod _{i\in I}f_{i}=f_{a}\cdot \prod _{i\in I\setminus \left\{ a\right\} }f_{i}\). By Proposition 1.181 (b) (the substitution rule for products: \(\left( u\cdot v\right) \circ g=\left( u\circ g\right) \cdot \left( v\circ g\right) \)), we have

By the induction hypothesis (applied to the set \(I\setminus \left\{ a\right\} \), which has size \(\left\vert I\right\vert -1\)),

Combining, we obtain

Let \(\left( f_{i}\right) _{i\in I}\in K\left[ \left[ x\right] \right] ^{I}\) be a multipliable family of FPSs. Let \(g\in K\left[ \left[ x\right] \right] \) be an FPS satisfying \(\left[ x^{0}\right] g=0\).

We shall first show the following auxiliary claim:

Claim 1: Let \(M\) be an \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\). Then, the set \(M\) determines the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\).

[Proof of Claim 1: The set \(M\) is an \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\). In other words, \(M\) is a finite subset of \(I\) that determines the first \(n+1\) coefficients in the product of \(\left( f_{i}\right) _{i\in I}\) (by the definition of “\(x^{n}\)-approximator”).

Let \(J\) be a finite subset of \(I\) satisfying \(M\subseteq J\subseteq I\). Then, Lemma A.43 (applied to \(J\) instead of \(I\)) yields

Also, Lemma A.43 (applied to \(M\) instead of \(I\)) yields

However, Proposition 1.364 (a) (applied to \(\mathbf{a}_{i}=f_{i}\)) yields

We also have \(g\overset {x^{n}}{\equiv }g\) (since the relation \(\overset {x^{n}}{\equiv }\) is an equivalence relation). Hence, Proposition 1.338 (applied to \(a=\prod _{i\in J}f_{i}\) and \(b=\prod _{i\in M}f_{i}\) and \(c=g\) and \(d=g\)) yields

In view of (1) and (2), this rewrites as

In other words,

(by the definition of the relation \(\overset {x^{n}}{\equiv }\)). Applying this to \(m=n\), we obtain

Forget that we fixed \(J\). We thus have shown that every finite subset \(J\) of \(I\) satisfying \(M\subseteq J\subseteq I\) satisfies

In other words, the set \(M\) determines the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\) (by the definition of what it means to “determine the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\)”). This proves Claim 1.]

Now, let \(n\in \mathbb {N}\). Lemma 1.363 (applied to \(\mathbf{a}_{i}=f_{i}\)) shows that there exists an \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\). Consider this \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\), and denote it by \(M\). Thus, \(M\) is an \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\); in other words, \(M\) is a finite subset of \(I\) that determines the first \(n+1\) coefficients in the product of \(\left( f_{i}\right) _{i\in I}\) (by the definition of “\(x^{n}\)-approximator”). Claim 1 shows that the set \(M\) determines the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\). Hence, there is a finite subset of \(I\) that determines the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\) (namely, \(M\)). In other words, the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\) is finitely determined.

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\), the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\) is finitely determined. In other words, each coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\) is finitely determined. In other words, the family \(\left( f_{i}\circ g\right) _{i\in I}\) is multipliable (by the definition of “multipliable”).

It remains to prove that \(\left( \prod _{i\in I}f_{i}\right) \circ g=\prod _{i\in I}\left( f_{i}\circ g\right) \).

In order to do so, we again fix \(n\in \mathbb {N}\). Lemma 1.363 (applied to \(\mathbf{a}_{i}=f_{i}\)) shows that there exists an \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\). Consider this \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\), and denote it by \(M\). Thus, \(M\) is an \(x^{n}\)-approximator for \(\left( f_{i}\right) _{i\in I}\); in other words, \(M\) is a finite subset of \(I\) that determines the first \(n+1\) coefficients in the product of \(\left( f_{i}\right) _{i\in I}\) (by the definition of “\(x^{n}\)-approximator”). Moreover, Proposition 1.364 (b) (applied to \(\mathbf{a}_{i}=f_{i}\)) yields

We also have \(g\overset {x^{n}}{\equiv }g\) (since the relation \(\overset {x^{n}}{\equiv }\) is an equivalence relation). Hence, Proposition 1.338 (applied to \(a=\prod _{i\in I}f_{i}\) and \(b=\prod _{i\in M}f_{i}\) and \(c=g\) and \(d=g\)) yields

However, Lemma A.43 (applied to \(M\) instead of \(I\)) yields

In view of this, we can rewrite (4) as

In other words,

(by the definition of the relation \(\overset {x^{n}}{\equiv }\)). Applying this to \(m=n\), we obtain

However, Claim 1 shows that the set \(M\) determines the \(x^{n}\)-coefficient in the product of \(\left( f_{i}\circ g\right) _{i\in I}\). Hence, the definition of the infinite product \(\prod _{i\in I}\left( f_{i}\circ g\right) \) (specifically, Definition 1.343 (b)) yields

Comparing this with (5), we obtain

Forget that we fixed \(n\). We thus have shown that each \(n\in \mathbb {N}\) satisfies

In other words, the FPSs \(\left( \prod _{i\in I}f_{i}\right) \circ g\) and \(\prod _{i\in I}\left( f_{i}\circ g\right) \) agree in all their coefficients. Hence, \(\left( \prod _{i\in I}f_{i}\right) \circ g=\prod _{i\in I}\left( f_{i}\circ g\right) \). This completes our proof of Proposition 1.399.