$ \newcommand{\hom}{\mathrm{hom}} $Suppose that $A, B$ are objects of a category with all finite products and exponentials, $\mathbb{C}$. Show that there is a bijection between $\hom(A, B)$ and $\hom(1, B^A)$ where $1$ is a terminal in $\mathbb{C}$.
My attempt: Consider the isomorphism given by transposition, $\hom(1 \times A, B) \cong \hom(1, B^A)$. It suffices to show that $\hom(A, B) \cong\hom(1 \times A, B)$. Since $1$ is terminal, there is unique morphism from any object $Z$ to $1$, $k_Z$. Consider $g \in \hom(A, B)$. We can define an isomorphism by $g \mapsto k_A \times g$. This must be an isomorphism since $k_A$ is unique.
This seems correct to me but I'm a little doubtful.
Your idea is right, but I don't think your proposed isomorphism $\newcommand\Hom{\operatorname{Hom}}\Hom(A,B)\simeq \Hom(1\times A,B)$ makes any sense at all.
I.e., you propose sending $g : A\to B$ to $k_A\times g : 1\times A \to A\times B$, which is not an element of $\Hom(1\times A,B)$.
Instead, it suffices to show that $A\simeq 1\times A$, since then $\Hom(A,B)\simeq \Hom(1\times A,B)$.
For this, we use the Yoneda lemma. Since we have $$\Hom(Z,1\times A)\simeq \Hom(Z,1)\times \Hom(Z,A) \simeq \{*\}\times \Hom(Z,A)\simeq \Hom(Z,A),$$ the Yoneda lemma tells us that $1\times A\simeq A$.
Edit
If you're not familiar with the Yoneda lemma, Matematleta points out that you can also see $A\simeq 1\times A$ by noticing that $(A,(k_A,\textrm{id}_A))$ is also a product for $1$ and $A$. (If $f:Z\to 1$ and $g:Z\to A$, then $g$ is the unique map from $Z$ to $A$ such that $k_A\circ g = f = k_Z$, and $\textrm{id}_A\circ g = g$).