Consider a Jordan matrix $\phi$ with the characteristic polynomial
$$\chi_\phi(t) = \prod_{i=1}^m(t - \lambda_i)^{n_i}$$
where $\lambda_i \ne \lambda_j$ for $i \ne j$. I want to show that $n_i$ is the sum of sizes of Jordan blocks with $\lambda_i$ on the diagonal.
It seems very logical, but I don't have an idea how to prove it and I don't find any proof on the net to help me.
The characteristic polynomial of a matrix $\phi$ is the determinant of the polynomial $t\mathbb-\phi$. (here, $t$ is supposed to be the diagonal matrix with $t$ on every diagonal entry).
A jordan matrix is triangular (upper or lower depends on convention), so $t-\phi$ is triangular as well.
The determinant of a triangular matrix is just the product of the diagonal entries.
Example with 3 jordan blocks: \begin{align} \phi&= \begin{pmatrix} 3 & & & \\ & 3 & 1 & \\ & & 3 & \\ & & & 5 \end{pmatrix} \\ t-\phi&= \begin{pmatrix} t-3 & & & \\ & t-3 & -1 & \\ & & t-3 & \\ & & & t-5 \end{pmatrix} \\ \mathrm{det}(t-\phi) &= (t-3)(t-3)(t-3)(t-5) \\ &=(t-3)^3(t-5) \end{align}
More abstractly: A Jordan matrix is a block matrix of the form: \begin{align} \phi=\begin{pmatrix} J_1 & & \\ & J_2 \\ &&... & \\ &&& J_n \end{pmatrix} \end{align} where each $J_k$ is of the form \begin{align} \begin{pmatrix} \lambda_k & 1 \\ & \lambda_k & 1 \\ & &... \\ &&&\lambda_k \end{pmatrix} \end{align}
then you can compute \begin{align} \mathrm{det}(t-\phi)&=\mathrm{det}\begin{pmatrix} t-J_1 & & \\ & t-J_2 \\ &&... & \\ &&& t-J_n \end{pmatrix} \\ &= \mathrm{det}(t-J_1)\mathrm{det}(t-J_2)...\mathrm{det}(t-J_n) \\ &= (t-\lambda_1)^{s_1}(t-\lambda_2)^{s_2}...(t-\lambda_n)^{s_n} \end{align} where $s_i$ are the sizes of the jordan blocks. Note that at this point the lambdas are not neccessarily all different. So you need to group them up, which means adding the exponents.