If $f:[0,\infty) \to \mathbb{R}$ be a differentiable function with $f(0)=1$ . . . Show that $f'(x)\geq e^x$ for all $x>0$.

Here is an answer for why it must be that $f(x)\geq e^x$:

This is in some sense a "reverse Grönwall inequality". Compute $g'(x)$, as suggested, using the product rule: $$g'(x)=e^{-x}(f'(x)-f(x))\geq 0.$$ Since $f(0)=1$, we have $g(0)=1$ as well, thus $g$ is a non-decreasing function with $g(0)=1$, implying that $g(x)\geq1$, so that $e^{-x}f(x)\geq1\Longrightarrow f(x)\geq e^x$. Thus $f'(x)\geq e^x$ as well.

(Edit: Remember that we assumed $f'(x)\geq f(x)$ to begin with, so $f(x)\geq e^x\Longrightarrow f'(x)\geq e^x$)

(2nd Edit: Also notice the $f(0)=1$ assumption is essential, as throwing it out yields the counterexample $f\equiv0$)

(3rd Edit: The relevance of $g$ is that we have shown $g(x)\geq 1$ for all $x$, which is equivalent the desired inequality)