Upper bound for $(1-1/x)^x$ I remember the bound $$\left(1-\frac1x\right)^x\leq e^{-1}$$ but I can't recall under which condition it holds, or how to prove it. Does it hold for all $x>0$?

Starting from $e^x \geq 1+x$ for all $x \in \mathbb{R}$:
For all $x \in \mathbb{R}$
$$ e^{-x} \geq 1-x. $$
For all $x \neq 0$
$$ e^{-1/x} \geq 1-\frac{1}{x}. $$
And, since $t \mapsto t^x$ is increasing on $[0,\infty)$, for $x \geq 1$
$$ e^{-1} \geq \left(1-\frac{1}{x}\right)^x. $$
