Let $T^{(i)}=A^{i-1}b$ for $i=1,\cdots n$. 
You can solve $\widehat{b}$ from the equation $T\widehat{b}=b$. 
By Cramer's rule, 

$$
\widehat{b_1}=\frac{det[b, T^{(2)}, \cdots, T^{(n)}]} {det T}
$$
$$
\widehat{b_2}=\frac{det[b, b, T^{(3)}, \cdots, T^{(n)}]} {det T}
$$
$\cdots$
$$
\widehat{b_n}=\frac{det[b, T^{(2)}, \cdots, T^{(n-1)},b]} {det T}
$$
Then you can see that 
$\widehat{b_1}=1, \widehat{b_i}=0$ for all $i=2,\cdots, n$.