# Characteristic of a field is not equal to zero

Let $$F$$ be a field such that for every $$x \in F$$ there exists a $$k >0$$ such that $$x^k=1$$. Does this imply that the characteristic of $$F$$ is strictly greater than zero?

Yes. If the characteristic is zero then the prime subfield is isomorphic to $$\mathbb{Q}$$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.
Every field $$F$$ of characteristic $$0$$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $$\frac{1}{2}$$, say, then there is no integer $$k>0$$ such that $$\left(\frac{1}{2}\right)^k = 1$$.
For a finite field with $$q$$ (prime power) elements, one has $$x^{q-1}=1$$ for each element $$x\ne 0$$ and $$x^q=x$$ for each element $$x$$.