Let $G$ be a connected matrix Lie group with Lie algebra $g$. Let $\Pi$ be a representation of $G$ acting on $V$ a finite dimensional vector space, and $\pi$ the associated representation of $g$. There is a theorem that says that in this case, $\Pi$ is irreducible iff $\pi$ is irreducible.
I am trying to understand the proof. I have understood the part where we show that an irreducible $\Pi$ implies an irreducible $\pi$. Nonetheless there is something I don't see in the proof of the converse.
The proof goes like this. Assume that $\pi$ is irreducible. Let $W$ be an invariant subspace of $V$. Then $W$ is invariant under $\Pi(e^{tX})$ where $t$ is a real number, and $X$ is an element of the Lie algebra $g$. The next claim is that this implies that $W$ is also invariant under $$\pi(X)=\frac{d}{dt}\Pi(e^{tX})\bigg|_{t=0}.$$ I don't see why. So, why does the fact that $W$ is invariant under $\Pi(e^{tX})$ make it also invariant under $\pi(X)=\frac{d}{dt}\Pi(e^{tX})\bigg|_{t=0}$?