Consider the autonomous ODE $$y^{(n)}+p_{1}y^{(n-1)}+\cdots+p_{n}y=0,$$ which has the characteristic equation $$P(r):=r^{n}+p_{1}r^{n-1}+\cdots+p_{n}=0.$$ Suppose that $$P(r)=(r-r_{1})^{m_{1}}(r-r_{2})^{m_{2}}\cdots(r-r_{k})^{m_{k}},$$ where $m_{1}+m_{2}+\cdots+m_{k}=n$. How can we show that the solutions $$t^{\ell}\mathrm{e}^{r_{i}t}\quad\text{for}\ \ell=0,1,\cdots,m_{i}-1\ \text{and}\ i=1,2,\cdots,k$$ are linearly independent?
When we do not have repeated roots, the proof follows from the Vandermonde determinant but I wonder to know if there is a generalization of it for the repeated case?
Consider the sum \begin{equation} \sum_{i=1}^{p}\sum_{\ell=0}^{m_{i}-1}c_{i\ell}\frac{t^{\ell}}{\ell!}\mathrm{e}^{\lambda_{i}t}=0,\tag{1} \end{equation} where $r_{i}\neq{}r_{j}$ if $i\neq{}j$. We claim that (1) implies $c_{i\ell}=0$ for $\ell=0,1,\cdots,m_{i}-1$ and $i=1,2,\cdots,p$. Assume the contrary. Without loss of generality, we may suppose that (1) can be written in the form \begin{equation} \sum_{i=1}^{q}\sum_{\ell=0}^{r_{i}}c_{i\ell}\frac{t^{\ell}}{\ell!}\mathrm{e}^{\lambda_{i}t}=0,\tag{2} \end{equation} where $q\leq{}p$, $r_{j}\leq{}m_{i-1}$ and $c_{ir_{i}}\neq0$ for $i=1,2,\cdots,q$. We may also suppose that $r_{i+1}\geq{}r_{i}$ for $i=1,2,\cdots,q-1$. We can write (2) in the form \begin{equation} \sum_{\ell=0}^{r_{1}}c_{1\ell}\frac{t^{\ell}}{\ell!} +\sum_{i=2}^{q}\sum_{\ell=0}^{r_{i}}c_{i\ell}\frac{t^{\ell}}{\ell!}\mathrm{e}^{(\lambda_{i}-\lambda_{1})t}=0.\tag{3} \end{equation} Differentiating (3) for a total of $(r_{1}+1)$-times, we get \begin{equation} \sum_{i=2}^{q}\sum_{\ell=0}^{r_{i}}c_{i\ell}\sum_{j=0}^{r_{i}+1}\binom{r_{i}+1}{j}\frac{t^{(\ell-j)}}{(\ell-j)!}(\lambda_{i}-\lambda_{1})^{r_{1}+1-j}\mathrm{e}^{(\lambda_{i}-\lambda_{1})t}=0\nonumber \end{equation} or equivalently \begin{equation} \sum_{i=2}^{q}\sum_{\ell=0}^{r_{i}}d_{i\ell}\frac{t^{\ell}}{\ell!}\mathrm{e}^{\lambda_{i}t}=0,\nonumber \end{equation} where for $i=2,3,\cdots,q$, $d_{i\ell}$ depends on $c_{i\ell},\cdots,c_{ir_{i}}$ such that $d_{ir_{i}}=(\lambda_{i}-\lambda_{1})^{r_{i}+1}c_{ir_{i}}\neq0$. Proceeding in this way, we see that \begin{equation} \sum_{\ell=0}^{r_{q}}e_{q\ell}\frac{t^{\ell}}{\ell!}\mathrm{e}^{\lambda_{q}t}\equiv0,\tag{4} \end{equation} where $e_{qr_{q}}=\Bigl[\prod_{j=1}^{q-1}(\lambda_{q}-\lambda_{j})^{r_{i}+1}\Bigr]c_{qr_{q}}\neq0$. Simply, (4) implies \begin{equation} \sum_{\ell=0}^{r_{q}}e_{q\ell}\frac{t^{\ell}}{\ell!}\equiv0.\nonumber \end{equation} Differentiating this for a total of $r_{q}$-times, we get $e_{qr_{q}}=0$. This is a contradiction.