Proof: By Induction
(related to Theorem: Bernoulli's Inequality)
We prove by induction that for all real \(x \ge 1\) and all natural numbers \(n\ge 2\), the following inequality holds:
\[(1+x)^n \ge 1 + nx\quad\quad ( * ).\]
 Base case $n=0$ the inequality \(( * )\) is fulfilled: $$(1+x)^0=1\ge 1+0x=1.$$
 Induction step $n\to n+1$
 Assume, the inequality \(( * )\) is fulfilled for all \(m \le n\).
 Because \(x \ge 1\), we have \(1 + x\ge 0\), and we can multiply the inequality \(( * )\) by \(1+x\) and get
\[\begin{align}(1+x)^{n+1}&\ge (1+nx)(1+x)\nonumber\\
&=1+(n+1)x+nx^2\nonumber\\
&\ge 1+(n+1)x.\nonumber\end{align}\]
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983