"I then did -21*row 4 + 6*row 3 to replace row 4 and got"
This is a determinant altering operation and not an elementary operation.
Don't write that $A$ equals something which isn't $A$.
Picking up where you errored and using the same idea you had one gets:
$$\begin{align} \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -6\cdot 16\\
0 & 0 & -21\cdot 6 & (-21)\cdot (-4)
\end{bmatrix}\\
&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}_.\end{align}$$
Making the proper compensation yields
$$\det(A)=-\dfrac{1\cdot 3\cdot (6\cdot 21)\cdot (-12)}{-21\cdot 6}=-36.$$
We can simply calculate the determinant of an opposite (lower) triangular matrix:
Let $J_n$ be the $n \times n$ matrix with $1$ on the anti-diagonal and $0$ otherwise (i.e. $J_ne_i = e_{n+1-i}$ for all $1 \leq i \leq n$, where $e_1, \dotsc, e_n$ denotes the standard basis). Given any $m \times n$-matrix $A$ the matrix $AJ_n$ originates from $A$ by vertically mirroring its colums from the middle, i.e. swapping the first column with the last, the second with the second last, etc.
If $A$ is an $n \times n$ square matrix then we get from $J_n^2 = I_n$ that
$$
\det(A) = \det(J_n) \det(AJ_n).
$$
In the case of $D_n$ we get that $D_n J_n$ is the $a_n$-scalar multiple a lower triangular matrix with diagonal entries $x-a_1, x-a_2, \dotsc, x-a_{n-1}, 1$, so
$$
\det(D_n)
= \det(J_n) \det(D_n J_n)
= \det(J_n) a_n (x-a_1) \dotsm (x-a_{n-1}).
$$
So the only difference is that we need to know $\det(J_n)$. Because $J_n$ is a permutation matrix, corresponding to $\sigma_n \in S_n$ with $\sigma(i) = n+1-i$, we have $\det(J_n) = \mathrm{sgn}(\sigma_n)$. Notice that
\begin{align*}
\sigma_{2n} &= (1 \;\; 2n) (2 \;\; n-1) \dotsm (n \;\; n+1) \\
\sigma_{2n+1} &= (1 \;\; 2n+1) (2 \;\; n-1) \dotsm (n \;\; n+2).
\end{align*}
So we can just count the number of transpositions used and get that
$$
\mathrm{sgn}(\sigma_n) =
\begin{cases}
\phantom{-}1 & \text{if $n \equiv 0,1 \bmod 4$}, \\
-1 & \text{if $n \equiv 2,3 \bmod 4$},
\end{cases}
= (-1)^{n(n-1)/2}.
$$
So alltogether we have
$$
\det(D_n) = (-1)^{n(n-1)/2} a_n (x-a_1) \dotsm (x-a_{n-1}).
$$
(The nice thing about this is that now that we have calculated $\det(J_n) = (-1)^{n(n-1)/2}$ we can use this to calculate the determinant of opposite triangular and opposite diagonal matrices more ore less in the usual way.)
Best Answer
Now you can expand by the first collumn, and you get
$$\begin{vmatrix} 0 & 0 &0 &-1 \\ 0 & 0 &-1 &-1 \\ 0 &-1 &-1 &-1 \\ -1 &-1&-1&-1 \end{vmatrix}$$
which is triangular, so the determinant equals $(-1)^4=1$.