Prove $\det(kA)=k^n\det A$

First, let's recall what multiplication of a matrix by a scalar means: When we multiply $\;kA$, where $A$ is an $n\times n$ matrix, and $k$ a scalar, then every entry $a_{ij}$ of matrix $A$ is multiplied by $k$: i.e. $a_{ij}\mapsto ka_{ij}$ for each $a_{ij}$. That means for each row $i,\;0\leq i\leq n,\;$ we can factor out $k$.

Now...Recall the elementary row operations you've learned, and how each one of them affects the determinant of the matrix on which it is operating. Specifically, when any one row is multiplied by the scalar $k$, the determinant of $A$ becomes $\;k\det A$. So given that scalar multiplication of a square $n\times n$ matrix is equivalent to "row operating" on $n$ rows (by multiplying each row by the scalar $k$), we can conclude that $$\large\det (kA) = \underbrace{k\cdot k\cdot\cdots \cdot k}_{\large n \;\text{times}}\det A = k^n\det A$$

Suppose that $v_1,\dots,v_n$ are the column vectors of the matrix $A$, then we can think of $\det$ instead of a function that maps one matrix to the reals as a function that maps $n$ vectors to the reals. Indeed, this is the way we define $\det$, it is a multilinear function of $n$ vectors! Now, being multilinear means that for each $i$ with $1\leq i\leq n$ we have:

$$\det(v_1\dots,kv_i,\dots,v_n)=k\det(v_1,\dots,v_i,\dots,v_n)$$

$$\det(v_1,\dots,u+w,\dots,v_n)=\det(v_1,\dots,u,\dots,v_n)+\det(v_1,\dots,w,\dots,v_n)$$

We simply need to use the first property. If $A=\begin{pmatrix}v_1 & \cdots & v_n\end{pmatrix}$ is your matrix, then by the usual definition of scalar multiplication for matrices we have $kA=\begin{pmatrix}kv_1 & \cdots & kv_n\end{pmatrix}$. In that case, we have that:

$$\det(kA)=\det(kv_1,\dots,kv_n)=k^n\det(v_1,\dots,v_n)=k^n\det(A)$$

Where $k^n$ appears because by multilinearity, for each index, one $k$ will be pulled out the $\det$. Pulling $n$ times the $k$ out, we get $k^n$ as desired.

Determinant is a multi-linear function, it means that $\det(x_1,\ldots,kx_i + y_i,\ldots,x_n) = k \det(x_1,\ldots, x_n) + \det(x_1,\ldots, y_i, \ldots,x_n)$, in which the i-th component is the i-th row of the matrix. by using $y_i = 0$ your desired result follows.