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
Please prove that:
  
  
*
  
*$ A = DP $ for a permutation matrix $ P $
  
*$ 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 $D^{T}$.
Could someone please advise me on how to solve this problem?
 A: 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!
A: 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{equation}
\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}
\end{equation}
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 $P$.
