Clarifying a proof for $G\mathbf{\pi}=0$

Question

While following a text on Continuous Time Markov Chains, I came across this proof for $G\mathbf{\pi}=0$. However, I fail to see how they get rid of the summation. Can anyone clarify why that relation is true?

Answer 1

Informally: equate like powers of $t$.

More formally: differentiate both sides $n$ times at $t=0$. The series converges exponentially fast (you may bound $\|\pi G^n\|$ by $\|\pi\| \|G\|^n$ where $\|G\|$ is the operator norm) and so the derivative may be moved inside the infinite sum.

Alternatively: $0$ is an analytic function, and so the only power series that converges to it at every point (note that the series on the right hand side does converge) is the series with all terms zero.