The claim results from the following statement:
(*) If $E/F$ is a degree $n$ separable extension, then the trace
of $E/F$ is nonzero.
To prove (*), denote by
the trace of $A$ over $K$, whenever $K$ is a field and $A$ a finite dimensional $K$-algebra.
t((K\otimes E)/K)=K\otimes t(E/K): K\otimes E\to K
for any extension $K/F$, the tensor products being taken over $F$, it suffices to check that $K/F$ can be chosen so that $t((K\otimes E)/K)\ne0$.
Thus, it only remains to observe that if $K/F$ splits the minimal polynomial of a generator of $E/F$, then the $K$-algebra $K\otimes E$ is isomorphic to $K^n$ by the Chinese Remainder Theorem.