Prove that the determinant of an upper triangular matrix is the product of its diagonal entries.
What I have so far:
We will prove this by induction for an $n$ $\times$ $n$ matrix. For the case of a $2 \times 2$ matrix, let A=$ \left( \begin{array}{ccc} a_{11} & a_{12} \\ 0 & a_{22} \end{array} \right)$. So det($A$)=$a_{11}a_{22}$ and the statement is true for the case of a $2 \times 2$ matrix. Now suppose that this statement is true for an $n$ $\times$ $n$ matrix. We will show that it also is true for an $(n + 1)$ $\times$ $(n + 1)$ matrix. Let A =$ \left( \begin{array}{ccc} a_{11} & a_{12} & \cdots & a_{1(n+1)}\\ 0 & a_{22} & \cdots & a_{2(n+1)} \\ \vdots & \cdots & & \vdots\\ 0 & 0 & \cdots & a_{(n+1)(n+1)}\end{array} \right)$.
I don't know what to do after this.