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Prove that for matrices $A, B\in M_n$ $$r(A)=r(B)=n \Rightarrow r(AB)=n$$

My attempt: $$\dim[\{A_1,...,A_n\}]=n$$ where $A_i$ are columns of $A$. $$\dim[\{B_1,...,B_n\}]=n$$ where $B_i$ are columns of $B$.

$$AB=C=\begin{pmatrix} \sum_{k=1}^{n} a_{1k}b_{k1} & \sum_{k=1}^{n}a_{1k}b_{k2} &\ldots &\sum_{k=1}^{n}a_{1k}b_{kn}\\ \vdots &\vdots &\ddots &\vdots\\ \sum_{k=1}^{n}a_{nk}b_{k1} &\sum_{k=1}^{n}a_{nk}b_{k2} &\ldots &\sum_{k=1}^n a_{nk}b_{kn}\\ \end{pmatrix}$$

$\dim[\{C_1,..,C_n\}]=?$ How do I continue from here?

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    Hint: $r(A)=n\iff\forall x\ne0\colon Ax\ne0$. – 2017-01-25

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$$r(A)=n\;\wedge\;r(B)=n\Rightarrow \det A\ne 0 \;\wedge\;\det B\ne 0$$ $$\Rightarrow \det (AB)=\det (A)\det (B)\ne 0 \Rightarrow r(AB)=n.$$

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    How does $\det (A)\det (B)\neq 0$ imply $r(AB)=n$? – 2017-01-25
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    It is a well kown result: A square matrix $M$ of order $n$ is non-singular iff its rank is equal to $n$. – 2017-01-25