$\overline{A}\simeq\overline{k}^n$ implies $A\simeq K_1\times\cdots\times K_r$

Let $A$ be a $k$-algebra which is finite as $k$-vector space. Denote by $\overline{A}:=A\otimes_k\overline{k}$, where $\overline{k}$ is the algebraic closure of $k$. I want prove $$\overline{A}\simeq\overline{k}^n \: \: \text{as }\overline{k} \text{-algebras}$$ if and only if $A$ is isomorphic as $k$-algebra to a product of finite separable extensions of $k$.

I'm not able to check that $\overline{A}\simeq\overline{k}^n$ implies that $A$ is isomorphic as $k$-algebra to a product of finite separable extensions of $k$. I'm only able to prove that $A$ is finite product of finite degree extensions of $k$, i.e. $A\simeq K_1\times\cdots\times K_r$ as ($k$-algebras), with $[K_i:k]<\infty$. Can you help me to check that every $K_i/k$ is separable?

A $k$-algebra $Q$ is separable iff for all algebraic extensions $K$ of $k$ the $K$-algebra $Q\otimes_kK$ is semisimple. If one of your $K_i$ were not separable, then for some finite extension $L$ of $k$ you'd have $K_i\otimes_kL$ not semisimple, so that $K_i\otimes_k\overline k$ would not be semisimple.