Let $K/k$ be a field extension of degree $n.$
If $K/k$ is separable. Then $K\otimes_{k} K \cong K^n \Leftrightarrow K/k$ is a normal extension.
I have a solution for $\Leftarrow$ direction:
By primitive element theorem there exists $\alpha \in K: K=k(\alpha).$ Let $f\in k[x]$ be the minimum polynomial of $\alpha$ over $k$. $$f(x)=\prod\limits_{i=1}^n \left(x-a_i\right).$$
Then $$K\otimes_{k} K \cong \frac{k[x]}{\langle f \rangle} \otimes_{k} k(\alpha) \cong \frac{k(\alpha)[x]}{\langle f\rangle}\cong \prod\limits_{i=1}^n \frac{k(\alpha)[x]}{\langle x-a_i\rangle} \cong K^n.$$
Can you please tell me why the $\Rightarrow$ direction is true?
Here is one idea. If $K/k$ is not normal, then the minimal polynomial $f(x)$ of $\alpha$ over $k$ won’t split into linear factors in $K$. This is because since $\alpha$ is a primitive element and is a root of $f$ in $K$, we know $f$ will not split into linear factors in any proper subfield $L$ with $k\subset L\subset K$; thus, if $f(x)$ does split into linear factors in $K$, then $K$ would be the splitting field of $f$ over $k$, making $K/k$ normal. So say instead that $f$ factors into irreducibles (in $K$) as $$ f(x) = (x-\alpha)\prod_{i=1}^N \pi_i(x) $$ where the $\pi_i(x)$ are distinct and at least one $\pi_i(x)$ has degree greater than $1$. (We know there are no repeated irreducible factors since $f(x)$ is separable and irreducible over $k$; if it had a repeated irreducible factor over $K$ then it would have repeated roots in an algebraic closure.)
Then using the same chain of isomorphisms, we have $$ K\otimes_k K\cong \frac{k[x]}{(f)}\otimes_k k(\alpha)\cong\frac{k(\alpha)[x]}{(f)}\cong K\times\prod_{i=1}^N \frac{k(\alpha)[x]}{(\pi_i(x))}. $$ And this is not isomorphic to $K^n$ as a $k$-algebra. There’s probably better ways of seeing this, but one way is to note that $K^n$ has $n$ maximal ideals whereas this product has fewer than $n$. To see this, recall that a maximal ideal in a direct product of rings $S=\prod_{i=1}^N R_i$ is of the form $$R_1\times\dots\times R_{j-1}\times M_j\times R_{j+1}\times\dots\times R_N, $$ where $M_j\subset R_j$ is a maximal ideal. (See this question for more details.) In other words, the number of maximal ideals in $S$ is equal to the sum of the number of maximal ideals in each of the $R_i$. Applying this to the direct product above, note that since one of the $\pi_i(x)$ has degree greater than $1$ and all the $\pi_i(x)$ multiply to give $f(x)$ (which has degree $n$), there are fewer than $n$ rings in the direct product including the $K$ out front. Moreover, each of those rings has a unique maximal ideal (the zero ideal, since they’re all fields). So there are strictly fewer than $n$ maximal ideals in that direct product, whereas $K^n$ has $n$ maximal ideals by the same reasoning.