Answer
See explanations.
Work Step by Step
Step 1. Prove the statement is true when $n=1$: $(1+x)^1\geq1+1\times x$ , thus it is true for $n=1$.
Step 2. Assume the statement is true when $n=k$: we have $(1+x)^k\geq 1+kx$, and with $x\gt -1$, we have $1+x \gt0$
Step 3. Prove it is true for $n=k+1$: we have (LHS stands for Left-Hand-Side)
$LHS-RHS=(1+x)^{k+1}-(1+(k+1)x)=(1+x)(1+x)^k-(1+kx+x)\geq (1+x)(1+kx)-(1+kx+x)=1+kx+x+kx^2-1-kx-x=kx^2\geq0$, thus we have $LHS\geq RHS$ which proves that the statement is also true for $n=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n$.
