Compute an $n\times n$ determinant with factorial and powers of $x$ Compute
$$
D_{n}=
\begin{vmatrix}
1 & 0 & 0 & 0 & 0 & \ldots & 1\\
1 & 1! & 0 & 0 & 0 & \ldots & x\\
1 & 2 & 2! & 0 & 0 & \ldots & x^{2}\\
1 & 3 & 3\cdot2 & 3! & 0 & \ldots & x^{3}\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ddots & \ldots\\
1 & n & n\left(  n-1\right)   & n\left(  n-1\right)  \left(  n-2\right)   &
n\left(  n-1\right)  \left(  n-2\right)  \left(  n-3\right)   & \ldots & x^{n}%
\end{vmatrix}
$$
My attempt: I tried to make an expansion along the first row, but I didn't
obtain anything; I also tried to compute some particular cases by I didn't see
how to obtain a general formula such that to be capable to compute $D_{n}$.
Any ideeas?

Let $M_n$ be the matrix of interest (the argument of the determinant in your question). We claim that 
$$M_n = A_n B_n, ~ A_n = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 1 & 1! & 0 & \cdots 
 & 0 \\ 1 & 2 & 2! & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & n & n(n - 1) & \cdots & n! \end{bmatrix}, ~ B_n = \begin{bmatrix} I_n & \vec{v}_{n-1} \\ 0 & (x - 1)^n / n!\end{bmatrix}$$
where $\vec{v}_{n-1}^T = \begin{bmatrix} 1 & x - 1 & (x - 1)^2 / 2 & \cdots & (x - 1)^{n-1} / (n-1)!\end{bmatrix}$. The correctness of the first $n$ columns of the product are trivial to verify. To verify the correctness of the last column, note that 
$$(A_n B_n)_{i, ~n+1} = \sum_{j = 0}^{i - 1} \frac{(i - 1)!}{(i - j - 1)!} \frac{(x - 1)^j}{j!} = \sum_{j = 0}^{i - 1} {i - 1 \choose j} (x - 1)^j = x^{i - 1}$$
by the Binomial Theorem. (Note here that the indexing of the matrix starts with   $1$, not zero.) As determinants are multiplicative, it follows that
$$\det(M_n) = \det(A_n) \cdot \det(B_n) = \left[ \prod_{k = 0}^n k! \right] \cdot (x - 1)^n / n! = \left[ \prod_{k = 0}^{n-1} k! \right] \cdot (x - 1)^n$$
since $\det(A_n)$ is simply the product of its diagonal entries (it's lower triangular) and $\det(B_n) = (x - 1)^n / n!$ by Laplace expanding along the last row. It's reassuring to see that this calculation confirms saulspatz's computer simulations. $\square$
