Your question is a great one, it is most welcome here! The answer is that, in a proof by induction, we first check the base case (here, it is $n=1$), and then, assuming the result is true for $n=k$, we prove that the result must also be true for $n=k+1$. In other words, we want to prove that $\text{true for }n=k\implies\text{true for }n=k+1$
Intuitively, this lets us say $\begin{align} (\text{base case}) \qquad\qquad\qquad\qquad\qquad\qquad&\text{true for }n=1\qquad\checkmark\\ {\text{true for }n=1,\text{ and }\atop (\text{true for }n=k\implies\text{ true for }n=k+1)}\bigg\}\implies&\text{true for }n=2\qquad\checkmark\\ {\text{true for }n=2,\text{ and }\atop (\text{true for }n=k\implies\text{ true for }n=k+1)}\bigg\}\implies&\text{true for }n=3\qquad\checkmark\\ \vdots\end{align}$
Thus, when we try to prove that the statement is true for $n=k+1$, i.e. $(1+x)^{k+1} ≥ 1 + (k + 1)x,$ we can use the assumption that the statement is true for $n=k$, i.e. $(1+x)^k ≥ 1 + kx.$ The reason why we have $(1+x)^k (1+x) ≥ (1+kx)(1+x)$ is that we are assuming $(1+x)^k ≥ 1 + kx$ is true, and then we multiply both sides by $(1+x)$.