Prove $ x^{\frac{1}{1-x}} < \frac{1}{e} $ for $ 0 \leq x < 1 $

Question

How can I prove the following statement?

If $ x \in \mathbb{R} $ and $ 0 \leq x < 1 $, then $$ x^{\frac{1}{1-x}} < \frac{1}{e}. $$

I could prove this statement: $ \lim\limits_{x \to 1} x^{\frac{1}{1-x}} = \frac{1}{e} $. I see that as $ x $ approaches $ 1 $, $ x^{\frac{1}{1-x}} $ approaches $ \frac{1}{e} $ but never exceeds $ \frac{1}{e} $. Now I am trying to come up with a proof that proves that it can never exceed $ \frac{1}{e} $.

Answer 1

One has $\log x < x - 1$. Note that $x-1< 0$, then $$\frac{1}{1-x}\log(x) < -1.$$

Answer 2

We need to prove that $$\frac{\ln{x}}{1-x}<-1$$ or $$x-1-\ln{x}>0.$$

Let $f(x)=x-1-\ln{x}$.

Thus, $$f'(x)=1-\frac{1}{x}=\frac{x-1}{x}<0$$ and since $\lim\limits_{x\rightarrow1}f(x)=0$, we are done!