I'm studying collocation mathods to solve ODEs and I got stuck on the proof of the Guillou-Soule theorem which proves that a collocation method is a particular kind of implicit RK method under some conditions on the coefficient of its tableau.
Disclaimer: In this question is not useful to know the backgrounds of ODEs methods nor anything else but polynomial interpolation. For those who are interested simply consider $u_h$ as a polynomial of degree $s$ and $\{c_1 \dots c_s\}$ are some points chosen in the interval $[0,1]$.
This is how starts the proof of Guillou-Soule's theorem given by Hairer at page 212 in his book "Solving Ordinary Differential Equations":
Interpolate with Lagrange polynomials $L_j (t) = \prod_{i \ne j} \frac{t - c_j}{c_i - c_j}$ the derivative of the collocation polynomial $\dot u_h$ in the points $\{t_0 + c_i h\}_{i = 1 \dots s}$ to get
\begin{equation} \tag{1} \label{1} \dot u_h(t_0 + t h) = \sum_{j = 1 \dots s} \dot u_h(t_0 + c_j h) L_j(t) \end{equation}
How this is true? I mean, the Lagrange interpolator in the nodes $\{\tau_i\}$ of a function $f$ is a polynomial $p \in \mathbb{P}_{s-1}$ in the form $$ p(t) = \sum_{j=1\dots s} f(\tau_j) L_j(t)$$
After this fact it seems that I have to prove $\dot u = p$, but anyways shouldn't \ref{1} be in the form $$ \dot u_h(t_0 + t h) = \sum_{j = 1 \dots s} \dot u_h(t_0 + c_j h) L_j(t_0 + t h) \qquad ? $$
Since $\;u_h(x)\;$ is a polynomial of degree $\,s,\,$ then the polynomial $\;u_h(t_0+th)\;$ is a polynomial of degree s by $\,t.\,$ In this way, the polynomial $\;\dot u_h(t_0+t h)=\dfrac{\partial}{\partial t}\,u_h(t_0+th)\;$ is a polynomial of degree $\;(s-1),\;$ which can be presented as a Lagrange interpolation polynomial with the nodes $\;t_0+c_1 h, t_0+c_2 h,\dots, t_0+c_s h,\;$ i.e. in the form $(1).$