I am self-studying out of McLarty's 'Elementary Categories, Elementary Toposes' and I am having trouble proving that if $\oslash$ is the intial object, and $A$ is any element in a cartesian closed category with $x: 1 \rightarrow A$ defined ($1$ the terminal element), then $\oslash^{A} \cong \oslash$. In particular, I am having trouble proving for any $B$, $\text{Hom}(\oslash^{A},B)$ has exactly one element. In the first part of the problem, we prove $\oslash^{A}$ is a subterminator, but I don't see how this is helpful for the second part of the problem.
Here's what I've tried so far: since $\oslash$ is initial, we always have $u: \oslash \rightarrow B$ for every $B$. Then we can exponentiate this map to get $u^{A}: \oslash^{A} \rightarrow B^{A}$ so that $u^{A}$ is the transpose of $u \circ ev: \oslash^{A} \times A \rightarrow B$. $u^{A}$ is therefore unique (I think). But now I don't see how this makes $\oslash^{A}$ initial (it doesn't seem to), since: 1) we haven't used the fact that $A$ has global elements; and 2) we would need to show that every object is of the form $B^{A}$ for this to make $\oslash^{A}$ initial.
This line of reasoning is making me think that I should be considering the problem from a different point of view. Another option I considered was trying to construct $\text{Hom}(\oslash^{A},\oslash)$, but I don't see how one can construct this other than by considering the map $ev: \oslash^{A} \times A \rightarrow \oslash$. I think if one shows this is unique then since $\oslash \rightarrow \oslash^{A} \times A$ exists and is unique, then $\oslash^{A} \times A \cong \oslash$ from which I think it follows that $\oslash^{A} \cong \oslash$, but I also don't see how this depends on $A$ having global elements. Intuitively, I can at least see that $A$ must at least not be initial, since it follows from the universal property of exponentiation that $\oslash^{\oslash} \cong 1$.
Thanks in advance for any help you can offer. Cheers.
Edit: I just wanted to mention an additional quick clarification for anyone else who might stumble across this post in the future. The accepted answer below is clear and useful, but I had a bit of trouble seeing the steps that fill in the argument regarding the functoriality of exponentiation--specifically the contravariant nature. McLarty actually gives a good intuitive discussion of this that I missed in my own hastiness. See section 6.5 in his book which discusses essentially the same fact but from a slightly different point of view. (He hasn't introduced functors yet at this point so he has to appeal to other means of arguing.)
In a cartesian closed category with an initial object, the initial object is strict, meaning that every morphism into it is an isomorphism. (This is a good exercise, but you can find a proof here.) Therefore, it suffices for your problem to exhibit some morphism $0^A \to 0$. However, if you have a morphism $a : 1 \to A$, then you can use functoriality of exponentiation to get a morphism $0^a : 0^A \to 0^1 \cong 0$. Consequently $0^A \cong 0$.