Answer
See below.
Work Step by Step
Proofs using mathematical induction consist of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number. Then we prove that then the statement also holds for $n + 1$.
Hence, here we have:
1) For $n=1: (1+x)^1=1+x\geq 1+x=1+(1)x$
2) Assume for $n=k: (1+x)^k\geq 1+kx$. Then for $n=k+1$:
$(1+x)^{k+1}=(1+x)(1+x)^k\geq(1+x)(1+kx)=1+(k+1)x+kx^2\geq1+(k+1)x$
Thus we proved what we wanted to.