Prove for all all $x\in\mathbb{R}: \exp(x-1) \geq x$ i just tried to solve this question, which is a small part of a bigger one.
Prove for all all $x \in \mathbb{R}: \exp(x-1) \geq x$
My first attempt was to simplify:
$$ e^{x-1} \geq x $$ $$ \Leftrightarrow \ln(e^{x-1}) \geq \ln(x) $$ $$ \Leftrightarrow x-1 \geq \ln(x) $$ $$ \Leftrightarrow x \geq \ln(x)+1 $$ $$ \Leftrightarrow e^x \geq e^{\ln(x)}+e^1 $$ $$ \Leftrightarrow e^x \geq x+e $$
My idea is that $e$ power $x$ is of course greater equal $x+e$. Is this correct  and if yes is it enough?

the answer Michael gave is a good answer but if you want different way,
$$e^{x-1}=x\implies x=1$$ now because both $x$ and $e^{x-1}$ are continuous you just need to check one value of $x$ that is above $1$ and one that is less. for example:$$\text{for $x<1,\,x=0$},\,e^{0-1}={1\over e}\gt0\\\text{for $x>1,\,x=2$},\,e^{2-1}=e\gt2$$
therefore $$e^{x-1}\ge1$$
