(This is taken direcly from the same exercise I've done quite a while ago, so it might seem a bit pedantic from time to time. In what follows $\langle b,c\rangle=(b,c)$) If $A$ is empty the identity is trivially true.
So we may assume $A$ is non-empty. We construct a bijection from $\left( A^{B} \right)^C$ to $A^{B\times C}$. Define $\psi$ by $\psi(f)=g$ where $f\in \left( A^{B} \right)^C$ and $g\in A^{B\times C}$ are such that
$$
g(\langle b,c\rangle) = \left( f(c) \right) (b) \;\;\; \text{for all $\langle b,c\rangle \in B\times C$}
$$
It is clear that $g$ is in $A^{B\times C}$ from its definition. Also, for each $f$ there exists unique $g=\psi(f)$ (since above definition of $g$ uniquely specifies its values on elements of $B\times C$) so $\psi$ is a function from $\left( A^{B} \right)^C$ to $A^{B\times C}$. We further claim that $\psi$ is in fact a bijection.
Injective: Suppose that $g=\psi(f_1)=\psi(f_2)$. Fix $c\in C$, then for every $b\in B$,
$$
\left( f_1(c) \right)(b)= g\left( \langle b, c\rangle \right)=\left( f_2(c) \right)(b)
$$
whence it follows that $f_1(c)=f_2(c)$. Since $c$ was arbitrary, then it follows that $f_1=f_2$, so $\psi$ is injective.
Surjective: Let $g\in A^{B\times C}$. Fix $c\in C$. Then define $f_c$ by
$$
f_c(b):=g\left( \langle b,c \rangle \right) \;\;\; \text{for all $b\in B$}
$$
Then clearly $f_c$ is a function from $B$ to $A$ (since $g$ is a function). Now define $f:C \to A^{B}$ by
$$
f(c)=f_c \;\;\; \text{for all $c\in C$}
$$
It follows that $\psi(f)=g$ by construction.
Thus $\psi$ is a bijection
