Proving $A^{B\cup C} \sim A^B\times A^C$

Question

let $A, B, C$ be sets so $B\cap C = \varnothing$. prove that $A^{B\cup C} \sim A^B\times A^C$.

I don't know how to start with solving this question. I know we need to define a function $A^{B\cup C} \to A^{B}\times A^{C}$ that is a bijective function but can't think of one. I'm looking for a simple function that can prove this theorem.

From Overview of basic results on cardinal arithmetic you can find these older questions: Let $A,B,C$ be sets, and $B \cap C=\emptyset$. Show $|A^{B \cup C}|=|A^B \times A^C|$ and Notation on proving injectivity of a function $f:A^{B;\cup; C}\to A^B\times A^C$. — Martin Sleziak, Dec 17 '17 at 11:39

José Carlos Santos · Answer 1 · 2017-12-17T15:15:10.190

Each function $f\colon B\cup C\longrightarrow A$ induces a function from $B$ to $A$ ($f|_B$) and a function from $C$ to $A$ ($f|_C$). This is always true. Now, If $f_B$ and $f_C$ are functions from $B$ into $A$ and from $C$ into $A$ respectively and if $B\cap C=\emptyset$, then$$\begin{array}{ccc}B\cup C&\longrightarrow&A\\ x&\mapsto&\begin{cases}f_B(x)&\text{ if }x\in B\\f_C(x)&\text{ if }x\in C\end{cases}\end{array}$$is a function from $B\cup C$ into $A$.

Proving $A^{B\cup C} \sim A^B\times A^C$

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