$\text{ Let }f\left(x\right)=(1+x{)}^n-(1+nx),x\in [-1,\mathrm{\infty }),\text{ then }\mathrm{f}$

$\begin{array}{l}f\left(x\right)=(1+x{)}^n-(1+nx)\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=n\left[(1+x{)}^{n-1}-1\right].\text{ Now for }\mathrm{x}=0,{\mathrm{f}}^{\mathrm{\prime }}\left(0\right)=0\text{ and for }\end{array}$

$\begin{array}{l}\mathrm{x}>0,{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)>0.\underset{\_}{\text{ Thus }}\mathrm{f}\text{ increases on }[0,\mathrm{\infty })\text{ i.e. }f\left(x\right)\ge f\left(0\right)\\ \text{ for }x\in [0,\mathrm{\infty })\text{. For }-1\le x<0,0\le 1+x<1\end{array}$

$\text{ so }(1+x{)}^n-1<0.\text{ So f decreases on }$

$\text{ [-1. 0) i.e. }f\left(x\right)\ge f\left(0\right)=0\text{ for }x\in [-1,0)\text{. Hence f has an }$$\text{absolute minimum at x = 0}$