Do representations of semisimple Lie algebras have any non-zero weights?

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I think I have reached an incorrect conclusion but need help finding the mistake in my logic.

Let $L$ be a semisimple Lie algebra. Then $D(L) = L$, where $D(L)$ is the derived subalgebra of $L$.

Suppose $\phi: L \rightarrow gl(V)$ is a finite dimensional representation. Then we have a decomposition of $V$ into subrepresentations $V_{\lambda}$, where $\lambda: L/D(L) \rightarrow gl(k)$ is a one dimensional representation of $L$, and $V_{\lambda} = \{v \in V: \forall x \in L, \exists n \in \mathbb{N}: (\phi(x) - \lambda(x))^n(v) = 0\}$. When a space $V_{\lambda}$ is non-zero we call $\lambda$ a weight.

Since $D(L) = L$, the only possible weights are $0$, since $L/D(L) \simeq 0$.

Is this reasoning correct?

This would further imply that for every $x \in L$, there exists $n \in \mathbb{N}$ such that $\phi(x)^n = 0$, using the fact that each $v \in V$ lies in $V_0$, and the fact that $V$ is finite dimensional.