OP is right: it’s enough to figure out a rather simple proof. This “rows based on a scalar multiple of the other” feature can be expressed in $A = C\,R$ form where $R$ is an 1 × n matrix (for example, any of non-zero rows of $A$) and $C$ is an n × 1 matrix. Then,
$$ A^2 = C\,R\,C\,R $$
(remember that matrix product is associative even for non-square matrices).
Look at the middle multiplication operation (of 3): what is “$R\,C$”? Matrix product of a row (at the left) and a column (at the right) gives 1 × 1 matrix. Then, let $c:=R\,C$, or more precisely, let c be the only element of said 1 × 1 matrix. (This kind of multiplication is known as the dot product.)
Obviously, matrix multiplication by an 1 × 1 matrix yields a scalar multiplication, so
$$ A^2 = C\,c\,R = c\cdot C\,R = c\cdot A$$