In a field, prove that if $ x + x = 0$ then $ 1 +1 = 0$

Question:

Suppose F is a field and that $ x$ cannot equal $ 0$. Prove that if $ x+x=0$ then $ 1 +1=0$.

My attempt:

We know $x^{-1}$ exists because $x$ cannot equal $0$.

$ x^{-1}\left(x+x\right)=\ x^{-1}.0$

$ x^{-1}.x + x^{-1}.x =0$, by distributivity.

$1 +1 = 0$, by multiplicative inverse.

I am not quite sure if this is the correct way to prove it. If it is incorrect could someone please tell me what I am doing wrong?

A solution that is a bit more general but based on the same idea, holds in any integral domain : if $x$ is non zero then it is regular and so $x+x =0 \implies x\cdot (1+1)=0$, and so by regularity, $1+1=0$

The real challenge here is to find a significant variant of Josh Mitkitzel's proof. But insofar as questions need answers, here's what popped into my head:

$1 + 1 = x^{-1}x + x^{-1}x = x^{-1}(x + x) = x^{-1}\cdot 0 = 0. \tag{1}$

A bona fide one-liner!