Solving $n$th order determinant I have a determinant of nth order that I am not able to convert into a triangular shape. I believe that this determinant is quite easy, but I can't find a way to fully convert one of the corners into zeros. My other idea was to use the Laplace principle, but that didn't work as well.
$$
        \begin{vmatrix}
        4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
        1 & 4 & 4 & 0 & \cdots & 0 & 0 \\
        0 & 1 & 4 & 4 & \cdots & 0 & 0 \\
        \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & 0 & 0 & \cdots & 4 & 4 \\
        0 & 0 & 0 & 0 & \cdots & 1 & 4 \\       
        \end{vmatrix}
$$
If someone could present a detailed way of converting this determinant into a triangular shape, it would be much appreciated. 
In addition to that, maybe someone could give some tips for solving nth order determinant by converting it into a triangular shape, using the Laplace principle or any other more or less basic methods.

Alternatively: doing row operations to get a diagonal matrix:
$$ R1\cdot (-\frac14)+R2\to R2; \ R2\cdot (-\frac13)+R3\to R3; \ etc$$
$$\begin{vmatrix}
        4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
        1 & 4 & 4 & 0 & \cdots & 0 & 0 \\
        0 & 1 & 4 & 4 & \cdots & 0 & 0 \\
        \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & 0 & 0 & \cdots & 4 & 4 \\
        0 & 0 & 0 & 0 & \cdots & 1 & 4 \\       
        \end{vmatrix}=\begin{vmatrix}
        4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
        0 & \frac{12}{4} & 4 & 0 & \cdots & 0 & 0 \\
        0 & 0 & \frac{32}{12} & 4 & \cdots & 0 & 0 \\
        \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & 0 & 0 & \cdots & \frac{n\cdot 2^{n-1}}{(n-1)\cdot 2^{n-2}} & 4 \\
        0 & 0 & 0 & 0 & \cdots & 0 & \frac{(n+1)\cdot 2^n}{n\cdot 2^{n-1}} \\       
        \end{vmatrix}= \\ \prod_{k=1}^n \frac{(k+1)\cdot 2^k}{k\cdot 2^{k-1}}=(n+1)\cdot 2^n.$$
