Eigenvalue of algebraic multiplicity $m$ is a pole of the resolvent of order $m$.

Question

Let $X$ be a Banach space and $T \in \mathcal{L}(X)$ be a bounded linear operator. Suppose that for some isolated point $\lambda \in \sigma(T)$ and some $m \in \mathbb{N}$ we have $\ker(T-\lambda I)^m = \ker (T-\lambda I)^{m+1}$. Is it true that then $\lambda$ is a pole of order $m$ of the resolvent $R_T(\mu) = (T-\mu I)^{-1}$?