If $x_1 , x_2,......x_n$ are non-zero elements of a field so is $\prod_{k=1}^n x_k$; and $\left(\prod_{k=1}^n x_k\right)^{-1} = \prod_{k=1}^n x_k^{-1}$.
Assume $n = 2$ true; How I did it: First: $$P(1) \implies \prod_{k=1}^1 x_k = x_1 non zero$$ from condition so $P(1)$ holds true.
For $n = 2$ also holds true from assumption so $x_1 * x_2$ non zero.
For $m = n-1$, assume $$P(m) \implies x_1 * x_2 * .....x_m$$ is non zero holds true.
$$P(m+1) \implies \prod_{k=1}^{m+1} x_k = \prod_{k=1}^m x_k * x_{m+1}$$ We have $$\prod_{k=1}^m x_k non 0$$ from $P(m)$ and since $x_{m+1}$ non zero from condition, from monotonic axiom of multiplication $P(m+1)$ is true if $P(m)$ is. Induction complete.
Next: $(\prod_{k=1}^n x_k)^{-1} = \prod_{k=1}^n x_k^{-1}$
$P(1)$ holds true: $$(\prod_{k=1}^1 x_k)^{-1} = (x_1)^{-1} = x_1^{-1} = \prod_{k=1}^1 x_k^{-1}$$ Also for $n = 2$ we assume it to be true: $$(\prod_{k=1}^2 x_k)^{-1 }= (x_1 * x_2)^{-1} = x_1^{-1} * x_2^{-1} = \prod_{k=1}^2 x_k^{-1}$$ For $m = n-1$, assume $P(m)$ holds true. $$P(m+1) = (\prod_{k=1}^{m+1} x_k)^{-1} = ((\prod_{k=1}^m x_k * x_{m+1})^{-1} = (\prod_{k=1}^m x_k)^{-1} * x_{m+1}^{-1} = \prod_{k=1}^m x_k^{-1} * x_{m+1}^{-1} = \prod_{k=1}^n x_k^{-1} \implies P(m+1)$$ holds true if $P(m)$ does.
It is hard to read the given argument. Roughly speaking, the idea looks ok but there is some confusion and it is unnecessarily complex. In particular, $x>0$ has no meaning in the context of a general field and I have never heard of the "monotonic axiom of multiplication".
To answer the question:
As here, we'll proceed by induction. For $n=1$ the claim is trivial. For $n=2$ we note that $a\neq 0$ implies that $a^{-1}$ exists so if we had $ab=0$ we could multiply by $a^{-1}$ to conclude that $b=0$. Thus, if $b\neq 0$ we must have $ab\neq 0$ Now, For a general $n$ note that, inductively, $x_1x_2\dots x_{n-1}\neq 0$ so we can apply the same argument as for $n=2$ to conclude that $x_1\dots x_n\neq 0$.
To handle the inverse case, note that $x_i\neq 0\implies x_i^{-1}\neq 0$ so we can simply apply the first case to $\{x_1^{-1},\dots,x_n^{-1}\}$