Proving that $ 1-u = e^{-u - \,u^2/2 - \,u^3/3 -...}$

Question

How can one see that for $-1 < u < 1$ we have the following equality $$ 1-u = e^{-u - \,u^2/2 - \,u^3/3 -...} \,\,\,\,?$$

It's probably easy to prove, however I've tried a couple of things so far (e.g. somehow using the series expansion of exp) but have failed.

Answer 1

this holds because for $-1\lt u \lt 1$, $-u-u^2/2-...=\ln(1-u)$ thus

$$e^{-u-u^2/2-...}=e^{\ln(1-u)}=1-u$$

Answer 2

For $|x|<1,$ using infinite Geometric Series,

$$(1-x)^{-1}=1+x+x^2+\cdots$$

Integrate either sides to find $$-\ln(1-x)=\sum_{r=1}^\infty\frac{x^r}r$$

Now $$\ln y=z\implies y=e^z$$

If $z$ is finite real, real $y>0$

Answer 3

Compute the Taylor expansion of the RHS: $$\begin{align} f(u)&=e^{-u-u^2/2-u^3/3-...},\\ f'(u)&=(-1-u-u^2-...)\ f(u)=\frac{f(u)}{u-1}\text{ (geometric series)},\\ f''(u)&=\frac{f'(u)(u-1)-f(u)}{(u-1)^2}=0\text{ (!)} \end{align}$$ Hence $$f(0)=1, f'(0)=-1, f''(0)=f'''(0)=...=0\implies f(u)=1-u.$$