analysis - Show that there exists a bijection from $(A^B)^C$ into $A^{B times C}$

Notation: Let A and B be sets. The set of all functions $f:A \rightarrow B$ is denoted by $B^A$.

Problem: Let A, B, and C be sets. Show that there exists a bijection from $(A^B)^C$ into $A^{B \times C}$. You should first construct a function and then prove that it is a bijection.

Actually this question hasn't been posted by me, but has already been answered and closed as Find a bijection from $(A^B)^C$ into $A^{B \times C}$

I don't agree, since this doesn't seen at least for me to be correct. Maybe I haven't got through the answer but in my view, the correct answer should be, following the same letters for the functions:

my Answer

Let $f \in (A^B)^C, g \in A^{B \times C}$. Define $\Phi: (A^B)^C \to A^{B \times C}$ by setting
$$\Phi(f)(b,c) = f(c)(b)$$

This is a bijection because it has an inverse $\Psi: A^{B \times C} \to (A^B)^C$

$$\Psi(g)(c)(b) = g(b,c)$$

I would like to know if my editions to the functions really answer the question or if the previous answer Find a bijection from $(A^B)^C$ into $A^{B \times C}$ was indeed correct.
Thanks.