Prove that $\frac{d^n}{dx^n}(1-e^x)^n\Big\rvert_{x=0}=(-1)^nn!$

Question

I'm not sure where to start with this. I have tried induction but I'm stuck on the inductive step. Could anyone let me know how to do this or give me a hint on how to approach this question?

I need to prove that the N'th derivative of $(1-e^x)^n=n!(-1)^n$ when evaluated at $x=0$

Answer 1

Look at the Maclaurin series: $$1-e^x=-x-\frac{x^2}2-\frac{x^3}{3!}-\cdots.$$ Therefore $$(1-e^x)^n=(-x)^n+\text{higher terms}.$$ Now differentiate $n$ times.

Answer 2

\begin{align*} \frac{d^n}{dx^n}(1-e^x)^n&=\frac{d^{n-1}}{dx^{n-1}}[-ne^x(1-e^x)^{n-1}]\\ &=\frac{d^{n-2}}{dx^{n-2}}[(-1)\cdot(-1)n(n-1)e^{2x}(1-e^x)^{n-2}+\cdots]\\ &\vdots\\ &=\frac{d^{n-(n-1)}}{dx^{n-(n-1)}}[(-1)^{n-1}n(n-1)\cdots (2)e^{(n-1)x}(1-e^x)+\cdots]\\ \frac{d^n}{dx^n}(1-e^x)^n\Bigg\rvert_{x=0}&=\left[(-1)^{n}n!e^{nx}+\underbrace{\cdots}_{\text{Terms having powers of }(1-e^x)}\right]\Bigg\rvert_{x=0}=(-1)^nn! \end{align*}