Let $k$ be a field. I would like to show that there are finitely many unital $k$-subalgebras of $k^n$. Let $s_n$ be the number of unital $k$-subalgebras of $k^n$. I would like to further show that $\sum \frac{s_n}{n!} x^n=e^{e^x-1}$.
Let $B$ be a unital subalgebra of $k^n$. Based on the formula of the generating series to show, it seems that the set of all $B$ is in bijection with the set of all partitions of $\{1,\dots,n\}$. For example, if $n=3$, then $B$ could (only) be of the form $\{(a,b,c):a,b,c\in k\}$, $\{(a,a,b):a,b\in k\}$, $\{(a,b,a):a,b\in k\}$, $\{(b,a,a):a,b\in k\}$, or $\{(a,a,a):a\in k\}$. But it seems quite difficult to prove, even whether $s_n$ is finite.
In fact, it is true that $s_n=B_n$, the $n$th Bell number (which is the number of equivalence relations on an $n$-element set).
Indeed, given any equivalence relation $\sim$ on $\{1,2,...,n\}$, one could define a unital subalgebra of $k^n$ consisting of all $n$-tuples $(a_1, a_2, ..., a_n)$ for which $a_i=a_j$ whenever $i \sim j$.
Conversely, given any unital subalgebra $B$ of $k^n$, one could define an equivalence relation $\sim_B$ where $i \sim_B j$ if and only if for all $(a_1, a_2, ..., a_n) \in B$, one has $a_i=a_j$.
To see that the above operations are inverse to each other, suppose that $\sim$ is an equivalence relation on $\{1,2,...,n\}$ and $B$ is its associated unital subalgebra. Then, it is clear that $\sim \subseteq \sim_B$ by the definition of $B$. On the other hand, if $i \not\sim j$, then the $n$-tuple with $1$ in the $k$th component for $k \sim i$ and $0$s in all the other components is in $B$, but the $i$th and $j$th components are not equal, so $i \not\sim_B j$. So, $\sim = \sim_B$.
Now, let $B$ be a unital subalgebra and $\sim_B$ be its associated equivalence relation. Also, let $B'$ be the unital subalgebra associated with $\sim_B$. Then, it is clear from the definitions that $B \subseteq B'$.
The harder direction is to show that $B' \subseteq B$ and hence $B=B'$. Since $B$ is a unital subalgebra of $k^n$, it contains $(a, a, ..., a)$ for all $a \in k$.
It suffices to show that the $n$-tuple with $1$ in the $j$th component for $j \sim_B i$ and $0$s in all the other components (henceforth to be denoted as $\mathbf{b}_i$) is in $B$ for all indices $i$, as the $\mathbf{b}_i$s are easily seen to generate $B'$ as a unital subalgebra. If $j \sim_B i$ for all indices $j$, then $\mathbf{b}_i=(1, 1, ..., 1)$, which of course is in $B$ since $B$ is a unital subalgebra. Otherwise, let $j$ be an index for which $j \not\sim_B i$. Then, by the definition of $\sim_B$, there is some $(a_1, a_2, ..., a_n) \in B$ for which $a_j \ne a_i$. Since $(a_j, a_j, ..., a_j) \in B$ by the above remark, one may subtract $a_j$ from all of the components and then multiply all of them by $(a_i-a_j)^{-1}$ to get an $n$-tuple in $B$ with the $i$th and $j$th components equal to $1$ and $0$ respectively. Next, repeat this process for the other indices $k$ with $k \not\sim_B i$. After that, the product of all of the resulting $n$-tuples would then be equal to $\mathbf{b}_i$, so $\mathbf{b}_i \in B$ since $B$ is a unital subalgebra.
This shows that $B' \subseteq B$ and hence $B=B'$.