I'm assuming (from Wikipedia) that the matrix is $\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & ... & a_n \\ b_1 & 0 & 0 & 0 & ... & 0 \\ 0 & b_2 & 0 & 0 & ... & 0 \\0 & 0 & b_3 & 0 & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & b_{n-1} & 0\end{pmatrix}$
Use induction on $n$. For $n = 1$ the proof is easy.
For $n > 1$ use Laplace expansion on the farthest right column to get
$\det \begin{pmatrix} t - a_1 & -a_2 & -a_3 & -a_4 & ... & -a_n \\ -b_1 & t & 0 & 0 & ... & 0 \\ 0 & -b_2 & t & 0 & ... & 0 \\0 & 0 & -b_3 & t & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & -b_{n-1} & t\end{pmatrix}$
= $t *\det \begin{pmatrix} t - a_1 & -a_2 & -a_3 & -a_4 & ... & -a_{n-1} \\ -b_1 & t & 0 & 0 & ... & 0 \\ 0 & -b_2 & t & 0 & ... & 0 \\0 & 0 & -b_3 & t & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & -b_{n-2} & t\end{pmatrix}$ $ + (-1)^n * \det \begin{pmatrix} -b_1 & t & 0 & ... & 0 \\ 0 & -b_2 & t & ... & 0 \\ 0 & 0 & -b_3 & ... & 0 \\ 0 & 0 & 0 & ... & t \\ 0 & 0 & 0 & ... & -b_{n-1} \end{pmatrix}$ $= t(t^{n-1} -a_1 t^{n-2} - a_2 b_1 t^{n-2} - ... - a_{n-1}b_1b_2...b_{n-2}) - b_1 b_2...b_{n-1}$ $= t^n - a_1 t^{n-1} - a_2 b_1 t^{n-2} - ... - a_n b_1 b_2 ... b_{n-1}$ as claimed.