I am trying to formally understand why $\|e^{A \tau}\|>1$, $\forall A \in\mathbb R^{n \times n}$, $\tau>0$ holds for every norm.
Attempt:
Proof A: Use matrix norm property: $\|MN\|\le \|M\|\|N\|$. Take $M:=e^{A\tau}, N:=e^{-A\tau}$. This gives $\|I\|\le\|e^{A\tau}\|\|e^{-A\tau}\|$. If $\|e^{A\tau}\|=\|e^{-A\tau}\|$ (which is not true however for any $A$), $\forall A \in\mathbb R^{n \times n}$
(Proof B: not totally sure how to do this, except to try Taylor series, may be?. I appreciate the suggestion here too, please). Then, we got $\|I\|\le\|e^{A\tau}\|^2 \Rightarrow \|e^{A\tau}\| \ge 1$. For $\tau>0$ and $A \ne 0$, we prove our claim.
I appreciate anyone clarifying with rigor both Proof A and Proof B.
Added: Can anyone show the detailed steps for at least the $2$-norm?
The conjecture is not true for a diagonal matrix $A=-3I$: $$ e^{t(-3I)} = e^{-3t}I \implies \|e^{tA}\|=e^{-3t} < 1 \mbox{ for all } t > 0. $$