Showing $1^A\cong 1$ in a CCC.

Here's the problem:

Let $\mathbb{C}$ be a cartesian closed category with terminal object $1$. Show that $1^A\cong 1$ for all objects $A$ of $\mathbb{C}$.

This is said to be trivial and I'm not surprised. However, I'm stuck. I don't want to use the Yoneda Lemma since the notes I'm using haven't introduced it. They assume familiarity with $\lambda$-calculus, too, which is new to me; I get the impression that it's strongly related (thanks to the "Curry-Howard Isomorphism"). I'm entirely self-taught in Category Theory.

Definition: A cartesian closed category (CCC) is a category $\mathbb{C}$ with binary products, exponentials, and terminal objects; that is, the following functors have right adjoints: $$! :\mathbb{C}\to 1,$$ $$\Delta :\mathbb{C}\to\mathbb{C}^{\cdot\,\cdot},$$ and $$\_\times A:\mathbb{C}\to\mathbb{C}$$ for all objects $A$ of $\mathbb{C}$.

The following 'deduction' is allowed: $$\frac{1\times A\cong A, f: A\to B}{\bar{f}:1\to B^A},$$

where $\bar{f}$ is "internal" to $B^A$.

My attempt: Since $1$ is terminal, there exist unique $t_A: A\to 1$, $t_{1^A}: 1^A\to 1$. Clearly $$\frac{1\times A\cong A, t_A: A\to 1}{\bar{t_A}:1\to 1^A}$$ by the above deduction. We also have $\mathbb{C}(1\times A, 1)\cong\mathbb{C}(1, 1^A)$ since $\_\times A\dashv (\_ )^A$. But $\lvert\mathbb{C}(1\times A, 1)\rvert =1$ since $1\times A\cong A$ and $t_A$ is unique so $\lvert\mathbb{C}(1, 1^A)\rvert =1$ too.

I doubt that what I've done so far is right. It's probably way off. What I want to show now is that the compositions of $\bar{t_A}$ and $t_{1^A}$ are the identity morphisms.

Hint. To show that an object is isomorphic to the terminal $1$, it is not required to exhibit the isomorphism. Remember : the terminal object is defined by a universal property up to (unique) isomorphism.
Meaning that you can just show $1^A$ to satisfy this universal property. I let you take it from here.
• Done it! Thank you, @Pece $\ddot\smile$ – Shaun Dec 7 '13 at 15:06