Let \(x\in\RR\).
If \(n\geq 0\) is an integer and \(x\geq -1\), then \((1+x)^n \geq 1 + nx\).
If \(n\geq 0\) is an even integer, then \((1+x)^n \geq 1 + nx\) for every \(x\in\RR\).
If \(r\geq 1\) is real and \(x\geq -1\), then \((1+x)^r \geq 1 + rx\).
By induction on \(n\): the case \(n = 0\) is \(1\geq 1\), and if \((1+x)^n\geq 1+nx\), then since \(1+x\geq 0\), \[ (1+x)^{n+1}\geq(1+nx)(1+x) = 1 + (n+1)x + nx^2\geq 1+(n+1)x. \]
For \(x\geq -1\) this is (1). For \(x<-1\) and even \(n\geq 2\), \((1+x)^n\geq 0\) while \(1+nx<1-n<0\); the case \(n=0\) is \(1\geq 1\).
The function \(\varphi(t)\coloneqq(1+t)^r\) is convex on \([-1,\infty)\) because \(\varphi''(t) = r(r-1)(1+t)^{r-2}\geq 0\) for \(t>-1\) and \(\varphi\) is continuous at \(-1\), so \(\varphi\) lies above its tangent line \(1+rt\) at \(t = 0\).
For all \(x\in\RR\), \[ 1-x \leq e^{-x} . \]