linear algebra - Proof clarification for inverse of matrix

This is from Artin Algebra, proposition 1.2.20:

Proposition: Let $A$ be a square matrix that has either a left inverse or a right inverse, a matrix $B$ such that either $BA=I$ or $AB=I$. Then $A$ is invertible and $B$ is its inverse.

Proof: Suppose $AB=I$. We perform row reduction on $A$. Say $A'=PA$, where $P=E_k\cdots E_1$ is the product of corresponding elementary matrices, and $A'$ is a row echelon matrix. Then $A'B=PAB=P$. Because $P$ is invertible, its bottom row is not zero. Then bottom row of $A'$ can't be zero either. $\dots$

I can't understand how bottom row of $A'$ can't be zero.

Answer

Suppose the bottom row of $A'$ is zero. Then, carrying out the matrix multiplication, you will see that the bottom row of $A'B$ is zero too.

If you're comfortable with block matrix partitioning, suppose

$$A' = \begin{bmatrix} A'_1 \\ 0\end{bmatrix}, B = \begin{bmatrix} B_1 & B_2 \end{bmatrix}$$

Then

$$A'B = \begin{bmatrix} A'_1 \\ 0\end{bmatrix}\begin{bmatrix} B_1 & B_2 \end{bmatrix} = \begin{bmatrix} A_1' B_1 & A_1' B_2 \\ 0 & 0 \end{bmatrix}$$

Which has a row of zeros.