Cartier dual of $\alpha_p$

How can I prove that the Cartier dual of $\alpha_p$ is again $\alpha_p$ (using the Yoneda lemma)? It should be something like $\alpha_p(R) \to (\alpha_p(R) \to \mu_p(R)), x \mapsto (y \mapsto \exp_{p-1}(x + y))$, where $\exp_{p-1}$ is the truncated exponential sequence. My problem is that this isn't a homomorphism.

Definition 1 Let $A$ be a finite commutative group scheme over $k$. Let $\hat{A}$ be the dual vector space to $A$. Define $\Delta : \hat{A} \to \hat{A} \otimes \hat{A}$ to be dual to the multiplication map $A \otimes A \to A$, and define a multiplication on $\hat{A}$ by the dual of the comultiplication of $A$. Then $\hat{A}$ is the Cartier dual.
Application to $\alpha_p$: As a ring, $\alpha_p$ is $k[t]/t^p$. Set $e_i = t^i$, for $0 \leq i \leq p-1$. In this basis, multiplication is given by $e_i e_j = \sum \delta_{i+j}^k e_k$ and comultiplication is given by $\Delta(e_k) = \Delta(t)^k = (t \otimes 1 + 1 \otimes t)^k = \sum \frac{k!}{i! j!} \delta_{i+j}^k e_i \otimes e_j$. (Here $\delta_a^b$ is $1$ if $a=b$ and $0$ otherwise.) Taking $f_i$ to be the basis of $\hat{A}$ with $\langle e_i, f_j\rangle = \delta_i^j i!$, we see that $A \cong \hat{A}$.
Definition 2: For any $k$-scheme $S$, the $S$-points of $\hat{A}$ are $\mathrm{Hom}(A \times S, \mathbb{G}_m \times S)$, with group structure given by multiplication of characters.
Application to $\alpha_p$: By a happy coincidence, I computed $\mathrm{Hom}(\mathbb{G}_a \times S, \mathbb{G}_m \times S)$ in an earlier answer. Every homorphism is of the form $t \mapsto e^{tN}$ for $N$ a nilpotent of $\mathcal{O}(S)$. If we want that morphism to factor through $\alpha_p$, a simple modification of that argument gives a bijection $$\mathrm{Hom}(\mathbb{G}_a \times S, \mathbb{G}_m \times S) \to \{ N \in \mathcal{O}(S) : N^p=0 \}.$$ with the bijection between the two sides again being that $N$ corresponds to the map $N \mapsto e^{tN}$.
Obviously, there is also a bijection between $\{ N \in \mathcal{O}(S) : N^p=0 \}$ and $\alpha_p(S)$. I leave it to you to check the necessary compatibilities for the full Yoneda proof.