We have f(x)=xn. So, f1(x)=nxn−1⇒f′(1)=n f2(x)=n(n−1)xn−2⇒f2(1)=n(n−1) f3(x)=n(n−1)(n−2)xn−3⇒f3(1)=n(n−1)(n−2) fn(x)=n(n−1)(n−3)…1⇒fn(1)=n(n−1)(n−3)…1 ⇒f(1)+1f1(1)+2!f2(1)+⋯+n!fn(1) =1+1n+2!n(n−1)+3!n(n−1)(n−2)+⋯+n!n(n−1)(n−2)…1 =nC0+nC1+nC2+⋯+nCn =2n