Prove that a square matrix can be expressed as a product of a diagonal and a permutation matrix.

I am having problems with this linear algebra proof:

Let $A$ be a square matrix of order $n$ that has exactly one nonzero entry in each row and each column. Let $D$ be the diagonal matrix whose $i^{th}$ diagonal entry is the nonzero entry in the $i^{th}$ row of $A$

For example:

$A = \begin{bmatrix}0 & 0 & a_1 & 0\\a_2 & 0 & 0 & 0\\0 & 0 & 0 & a_3 \\0 & a_4 & 0 & 0 \end{bmatrix} \quad$ $D = \begin{bmatrix}a_1 & 0 & 0 & 0\\0 & a_2 & 0 & 0\\0 & 0 & a_3 & 0\\0 & 0 & 0 & a_4 \end{bmatrix}$

A permutation matrix, P, is defined as a square matrix that has exactly one 1 in each row and each column

1. $A = DP$ for a permutation matrix $P$
2. $A^{-1} = A^{T}D^{-2}$

My attempt:

For 1, I tried multiplying elementary matrices to $D$ to transform it into $A$:

$$A = D * E_1 * E_2 * \cdots * E_k$$

Since I am performing post multiplication with elementary matrices, the effect would be a column wise operation on D. But I can't see how this swaps the elements of $D$ to form $A$. I also cannot prove that the product of the elementary matrices will be a permutation matrix.

For 2, my attempt is as follows (using a hint that $PP^{T} = I$):

\begin{aligned} A^{T}D^{-2} &= (DP)^{T}D^{-2} \\ &= (P^{T})(D^{T})(D^{-1})(D^{-1}) \\ &= (P^{-1})(D^{T})(D^{-1})(D^{-1}) \end{aligned}

I am not sure how to complete the proof since I cannot get rid of the term $D^{T}$.

Hint: For $(1)$, find a matrix $P(i,j)$ that swaps columns $i$ and $j$. Your permutation matrix will be a product of $P(i,j)$'s.
For $(2)$, try to convince yourself that when $D$ is diagonal, $D^{T}=D$. It's not too hard!
For part one, you have the right idea. Try multiplying $A$ on the right by various permutation matrices, and just see what happens. For example, multiplying on the right by this matrix swaps columns 1 and 3. $$\begin{bmatrix} 0 & 0 & a_1 & 0 \\ a_2 & 0 & 0 & 0 \\ 0 & 0 & 0 & a_3 \\ 0 & a_4 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} a_1 & 0 & 0 & 0 \\ 0 & 0 & a_2 & 0 \\ 0 & 0 & 0 & a_3 \\ 0 & a_4 & 0 & 0 \end{bmatrix}$$ You can find similar matrices to do more column swaps to get to $D$. You don't need to prove in general that these column swap matrices are necessarily permutation matrices (although they are), since all this problem asks you to do is write down a matrix $P$.