Why a root $\alpha_i$ with multiplicity $m_i$ will satisfy up to $(m_i-1)$-th derivatives of polynomial of degree $n$? I'm reading a proof of linear recurrence relation with constant coefficients of order $k$, which gives the formula of $a_n$ when some roots of its characteristic equation has multiplicity $\ge2$ . In one step it says:
\begin{align*}
&\textrm{Since }\alpha_i\textrm{ satisfy}\\
&C_n\alpha^n+C_{n-1}\alpha^{n-1}+\dots+C_{n-k}\alpha^{n-k}=0,\,\,\,\,\,\,\,\textrm{(A)}\\
&\textrm{and }\alpha_i\textrm{ has multiplicity }m_i,\textrm{it satisfies up to }(m_i-1)\textrm{-th derivative of (A).} 
\end{align*}
So what's the reason the last sentence makes sense?

The theorem:
\begin{align*}
&\textrm{Let }k\in\mathbb Z^+, C_n,C_{n-1},\cdots,C_{n-k}\in\mathbb R,\textrm{and }C_n,C_{n-k}\not=0;\\
&C_na_n+C_{n-1}a_{n-1}+\dots+C_{n-k}a_{n-k}=f(n)\\
&\textrm{Assume }\alpha_1,\alpha_2,\dots,\alpha_t\textrm{ are its characteristic roots, and }\alpha_i\textrm{ has multiplicity }m_i,\\
& 1\le i\le t\textrm{ and let }u_i(n)=(d_{i_0}+d_{i_1}n+\dots+d_{i_{m_i-1}}n^{m_i-1})\alpha_i^n,\textrm{where }d_k\textrm{ are any constant, then}\\
&a_n=u_1(n)+u_2(n)+\dots+u_t(n).
\end{align*}
 A: Let r be a root of p(x) in R[x] with multiplicity k.
Then p(x) = (x - r)$^k$ q(x) for some q(x) in R[x].
By induction show r is a root of the j-th derivative
for all j's upto k - 1.
A: If $\alpha_i$ is a root of
$p(x) = 0 \tag 1$
of multiplicity $m_i \ge 2$, then 
$p(x) = (x - \alpha_i)^{m_i} q(x), \tag 2$
where $q(x)$ is some polynomial with
$\deg q = \deg p - m_i; \tag 3$
we see that
$p'(x) = m_i(x - \alpha_i)^{m_i - 1}q(x) + (x - \alpha_i)^{m_i} q'(x), \tag 4$
whence
$p'(\alpha_i) = m_i(\alpha_i - \alpha_i)^{m_i - 1}q(x) + (\alpha_i - \alpha_i)^{m_i} q'(x) = 0, \tag 5$
and $\alpha_i$ is a root of $p'(x)$; we observe that (4) may be written
$p'(x) = (x - \alpha_i)^{m_i - 1} (m_iq(x) + (x - \alpha_i)q'(x)), \tag 6$
which is in the same form as (2) but with $m_i$ replaced by $m_i - 1$; the similarity 'twixt (2) and (6) suggests a simple induction:  taking $m_i = 2$ as the base case, we assume that $m_i = k$, and that for any $p(x)$ such that
$p(x) = (x - \alpha_i)^k q(x) \tag 7$
$\alpha_i$ is a root of both $p(x)$ and it's first $k - 1$ derivatives:
$p^{(j)}(\alpha_i) = 0, \; 0 \le j \le k - 1; \tag 8$
then if
$p_1(x) = (x - \alpha_i)^{k + 1} q_1(x), \tag 9$
as in (6) we have
$p_1'(x) = (x - \alpha_i)^k ((k + 1)q_1(x) + (x - \alpha_i)q_1'(x)); \tag{10}$
clearly $\alpha_i$ is a zero of $p_1(x)$, and applying our inductive hypothesis to (10) we affirm that $\alpha_i$ is a root of both $p_1'(x)$ and its first $k - 1$ derivatives, that is, $\alpha$ satisfies $p_1(x)$ and the first $k$ derivatives thereof, the precise affirmation we seek.  Thus we have inductively demonstrated that for all $m_i$,
$p(x) = (x - \alpha_i)^{m_i}q(x) \Longrightarrow p^{(j)}(\alpha_i) = 0, \; 0 \le j \le m_i - 1. \tag{11}$
