$ABC=I\implies B$ is invertible and $B^{-1} = CA$ $A$, $B$ and $C$ are square matrices with $ABC=I$. I need to show that  $B$ is invertible and $B^{-1} = CA$.
I have proved it using the fact stated here.
Since we only need to prove invertiblity of $B$ there may be a simpler proof. Any ideas? 
Here is my proof.
$(AB)C=I \implies C(AB) = I \implies (CA)B=I \implies B^{-1}=CA$
Here I have used the fact in linked question. I am seeking an elementary proof but seems like there isn't one. 
 A: All matrices are invertible. The we obtain:
$ABC=I \iff AB=C^{-1} \iff A=C^{-1}B^{-1} \iff CA=B^{-1}$
A: Using the fact mentioned in your question we see that $A$ and $C$ are invertible and so $B$ is also invertible and that
$$B=A^{-1}C^{-1}$$
Can you take it from here?
A: First of all, it's clear that $A$ is invertible, because it has the inverse $BC$.
If you want to know that $B^{-1}=CA$, try writing their product and showing that it must be $I$:
$B(CA) = IBCA = A^{-1}ABCA=A^{-1}(ABC)A = A^{-1}IA = I$
A: As stated by Balla, by properties of the determinant , $$Det(ABC)=DetADetBDetC=1 $$, so each of DetA, DetB, DetC must be nonzero, so invertible. You can then check that $ C^{-1}B^{-1}A^{-1}$ is the inverse of $ABC$, since : $$(C^{-1}B^{-1}A^{-1})(ABC)=C^{-1}B^{-1}(AA^{-1})BC=....=I$$, so that $$B^{-1}C^{-1}A^{-1}=I $$ Now you can multiply both sides by $AC$, to find the inverse of $B$.
A: As many answers mention, it is fairly easy to show that $A$ (or $C$, if you prefer) must be invertible.


*

*We can use the properties of determinants: $\det(A) \det(B) \det(C) = \det(ABC) = \det(I) = 1$, so $A$ must be invertible.

*We can use the rank inequality $\text{rank}(PQ) \leq \min\{\text{rank}(P),\text{rank}(Q)\}$, setting $A = P$ and $BC = Q$, noting $PQ = I$ has full rank


With that in mind, we have the following proof of a slightly different flavor.
$$
B(CA) = A^{-1}(ABC)A = A^{-1} I A = I
$$
So that $CA = B^{-1}.$
