Prove that $$\lfloor e^x \rfloor =\lfloor e^{\lfloor x \rfloor} \rfloor \tag{1}$$
I was actually trying to prove $$\lfloor \sqrt{x} \rfloor=\lfloor \sqrt{\lfloor x \rfloor} \rfloor$$ and i succeeded since i started with:
$$\lfloor \sqrt{x} \rfloor =m$$ and I have seen a generalized result:
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous increasing function , Then $\lfloor f(\lfloor x\rfloor)\rfloor=\lfloor f(x)\rfloor.$
In similar manner i started trying to prove $(1)$
$$\lfloor e^x \rfloor=m$$ where $m \in \mathbb{Z}$
Case $1.$ If $x \lt 0$, Then we have $0 \lt e^x \lt 1$ Hence $$\lfloor e^x \rfloor=0$$
Also when $x \lt 0$, We have $\lfloor x \rfloor \lt 0$ $\implies$ $0 \lt e^{\lfloor x \rfloor} \lt 1$
Hence $$\lfloor e^{\lfloor x \rfloor} \rfloor=0$$
Case $2.$ If $x \ge 0$, Then $m \ge 1$
Letting $$\lfloor e^x \rfloor=m$$ We have
$$m \le e^x \lt m+1$$ $\implies$
$$\ln(m) \le x \lt \ln(m+1)$$
How to proceed from here?