I came across this proof on p.86 of Elementary Linear Algebra by Howard Anton regarding the subject line.
$kA$ is invertible for any non-zero scalar $k$ and $(kA)^{-1} = k^{-1}A^{-1}$.
$(kA)(k^{-1}A^{-1})=k^{-1}(kA)A^{-1}=(k^{-1}k)AA^{-1}=(1)I =I$ similarly $(k^{-1}A^{-1})=(kA)=I$. Thus $kA$ is invertible and $(kA^{-1}) = k^{-1}A^{-1}$.
I am confused of the part right after "similarly". Shouldn't $(k^{-1}A^{-1}) = (kA)^{-1}$? That is what it says at the end of the first line.
Obviously you certainly do not have $k^{-1}A^{-1} = kA$, take for example $A=I$ which would mean that basically $k^{-1}=k$ for any non-zero $k$.
The parenthesis around $kA$ doesn't make sense unless you actually invert it. Unfortunately there's a parenthesis around $k^{-1}A^{-1}$ which doesn't make sense either.
What it probably should read is that $(k^{-1})A^{-1})(kA) = I$ that way the parenthesis make some sense (although multiplication is associative). Also it is similar to the first equality.
Note that it would be erroneous to write $(kA)^{-1}$ before you actually established that it is invertible, that is, $k^{-1}A^{-1}$ is the inverse (which is done with the two parts of the proof).