In an algebraically closed field, does every nonzero element have n distinct n-th roots?

Question

Let $F$ be an algebraically closed field. This certainly implies that every non-zero element $x$ of $F$ has at least one $n$-th root, for each positive integer $n$. Does it in fact imply the condition that every non-zero element has $n$ distinct $n$-th roots?

Answer 1

Hint: Recall that if $F$ is a field of characteristic $p,$ then $(a + b)^p = a^p + b^p.$ How many distinct $p$th roots of unity does this imply are there in $F$?

Hint: What equation must a $p$th root of unity satisfy? Once you have this, apply the first hint.