Prove the determinant is the product of its diagonal entries

Using the cofactor expansion along the first column we get $$|A|=(-1)^{1+1}a_{11} \left |\begin{matrix} a_{22}&a_{23} & \cdots & a_{2(n+1)} \\ 0 &a_{33} & \cdots & a_{3(n+1)} \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & a_{n+1n+1} \end{matrix}\right |$$ (Note that remaining terms in the expansion are zero.)

Using the induction hypothesis (you know det of the $n\times n$ matrix), we get

$|A|$=$a_{11}a_{22}a_{33}\cdots a_{n+1n+1}$

Use the Laplace expansion to finish the induction. If you take the minor $A_{12}$ for example, you'll get the determinant of an upper triangle matrix with top-left entry $0$, which by induction is $0$. The only minor of the top row which is not $0$ is $A_{11}$.

[edit] Used the notation for minors found here.