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

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 $x=0$

Sameer Baheti · Answer

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)^nBiggrvert_{x=0}&=left[(-1)^{n}n!e^{nx}+underbrace{cdots}_{text{Terms having powers of }(1-e^x)}right]Biggrvert_{x=0}=(-1)^nn!
end{align*}

Angina Seng · Answer

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.

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

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