Show that for $0<||A||<\epsilon$ we have $||e^{A}-I||>0$

Question

I want to show that there exists $\epsilon>0$ such that if an $n \times n$ matrix $A$ satisfies $0<||A||<\epsilon$, we have $||e^{A}-I||>0$.

This seems pretty simple, but my attempts to derive a strict lower bound haven't gone anywhere. I have tried expanding out the matrix power series and using the norm triangle inequality but that hasn't been very fruitful. Some help would be much appreciated.

Answer 1

Note that $$\begin{align*} ||e^A-I|| &= \left|\left|A + \sum_{n=2}^\infty \frac{A^n}{n!}\right|\right| \\ &\geq ||A|| - \left|\left|\sum_{n=2}^\infty \frac{A^n}{n!}\right|\right| \\ &\geq ||A|| - \sum_{n=2}^\infty\left|\left|\frac{A^n}{n!}\right|\right| \\ &\geq ||A|| - \sum_{n=2}^\infty \frac{||A||^n}{n!} \\ &= 1 + 2||A|| - e^{||A||} \end{align*}$$

Now just show that $1+2x-e^x > 0$ for small $x>0$.