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.