Irreducible components of hypersurface This is exercice 5.4 from Miles Reids Undergraduate Commutative Algebra.
Let $k$ be an algebraically closed field. For $f \in k[x_1,...,x_n]$, write $V(f) \subset k^n$ for the hypersurface defined by $f=0$. 
Using Hilberts Nullstellensatz, one can show that:
Claim: if $f$ is irreducible and $f$ does not divide $g$, then $V(f) \not\subset V(g)$ (see this post of mine).
From this result, I want to deduce the following.
If $g= $  cst $\cdot \, \,\Pi \, \,f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. 
If I understand correctly, this follows from the counterpositive statement of the claim above. 
Indeed, for any $f_i$, we have that $V(f_i) \subset V(g)$ and hence $f_i$ is either irreducible or $f_i$ divides $g$. If $f_i$ is irreducible, we are done. If not, we can reiterate the process until we get irreducibility. 
Does this make sense? 
Thank you for all your comments and remarks. 
 A: You can't just use the contrapositive of the quoted result, you need the if and only if form. I.e., that $f\mid g$ if and only if $V(f)\subseteq V(g)$.
Your proof is phrased rather weirdly, and I can't quite tell what you're saying. I thought that the $f_i$s were the irreducible factors of $g$, but then you conclude that $f_i$ is either irreducible or $f_i$ divides $g$, which makes no sense as a conclusion if the $f_i$s are irreducible factors, since we know both of those things about $f_i$ from the start.
The correct proof goes like this.
Let $g=c\prod_i f_i^{n_i}$ with the $f_i$s irreducible polynomials with $f_i$ not an associate of $f_j$ for $i\ne j$, and $c\in k$.
Then $f_i\mid g$, so $V(f_i)\subseteq V(g)$. Moreover, if $a\in k^n$ with $g(a)=0$, then $\prod_i f_i(a)=0$, so for some $i$, $f_i(a)=0$. Thus $a\in V(f_i)$ for some $i$. Thus $V(g)\subseteq \bigcup_i V(f_i)$. Thus we have that
$$ V(g) = \bigcup_i V(f_i).$$
It just remains to check that $V(f_i)$ is irreducible. Suppose not, then there are $g$ and $h$ such that $V(f)\subseteq V(gh)$, but $V(f)\not\subseteq V(g)$ and $V(f)\not\subseteq V(h)$. Using that $f\mid g$ if and only if $V(f)\subseteq V(g)$, this translates to the existence of $g$ and $h$ such that $f_i\mid gh$, but $f_i\nmid g$ and $f_i\nmid h$, which contradicts the irreducibility of $f_i$, since irreducibles are prime in $k[x_1,\ldots,x_n]$.
