Exponential objects in a cartesian closed category: $a^1 \cong a$

For any object $x$, we have: $$\operatorname{Hom}(x,a^1)\cong \operatorname{Hom}(x\times 1,a)\cong \operatorname{Hom}(x,a)$$ So Yoneda's lemma gives us that $a$ is isomorphic to $a^1$.

You can also reason as follows, without the Yoneda lemma. But proving uniqueness of right adjoints is cumbersome without using Yoneda, and easy with. Anyway, here it goes:

The functor $(-)\times 1$ is isomorphic to the identity functor. The identity functor is a right adjoint of itself, so the identity functor is also right adjoint to $(-)\times 1$. Then uniqueness of right adjoints gives that $(-)^1$ is isomorphic to the identity functor.

I know that this is a bit late, but I'm readong the book right now and found this post, so i'll give my solution anyway, which doesn't use the Yoneda Lemma, or adjunctions, etc.

The idea is to notice that $id_a : a\to a$ is a terminal object in the comma category $C\to a$. But the definition of the exponential allows you to show that $ev : a^1 × 1 \to a$ is also a terminal object, and therefore isomorphic to the first one. It is then easy to conclude that $a^1×1 \simeq a$ and so $a^1 \simeq a$