Derivation of Binomial theorem

Question

Let $F(x)$ be the function defined by (1):

$$ 1)F(x)=(1-e^{-ax})^N$$ Using the binomial theorem $F(x)$ can be written as (2):
cc\begin{equation} 2)F(x)=\sum_{n=0}^{N}\binom{N}{n}(-1)^ne^{-axn} \end{equation}

Now the Derivation of $F(x)$ is $f(x)=F'(x)$ $$ 3) f(x)=aNe^{-ax}(1-e^{-ax})^{N-1}$$ Similar using binomial theorem $$ 4)f(x)=aN\sum_{n=0}^{N-1}\binom{N-1}{n}(-1)^ne^{-ax(n+1)}$$

My question how we derive (2) to get (4)?

Answer 1

$$\left(\sum_{n=0}^{N} \binom{N}{n} (-1)^n e^{-axn}\right)' = \sum_{n=1}^{N} an\binom{N}{n} (-1)^{n+1} e^{-axn}$$ then we use the fact $$k \binom{n}{k} = n\binom{n-1}{k-1}$$ so $$\sum_{n=1}^{N} an\binom{N}{n} (-1)^{n+1} e^{-axn} = \sum_{n=1}^{N} aN\binom{N-1}{n-1} (-1)^{n+1} e^{-axn} = \sum_{n=0}^{N-1}aN\binom{N-1}{n}(-1)^{n}e^{-ax(n+1)}$$

Answer 2

Hint: After derivation use the binomial identity $\binom{N}{n}=\frac{N}{n}\binom{N-1}{n-1}$ and then apply an index shift.