Well known facts in extreme value theory:
Let $\{X_i\}_{\forall i \in \{1,...,n\}}$ be i.i.d. random variables with cdf $F$. If there exists $\{a_n\}_{n\in \mathbb{N}}>0$, and $\{b_n\}_{n\in \mathbb{N}}\in \mathbb{R}$ such that $Z_n\equiv \frac{M_n-b_n}{a_n} \Rightarrow_n Z$, where $Z$ has distribution of the same type as Gumbel and $M_n\equiv \max_{i\in \{1,...,n\}}X_i$, then we say that $F$ is in the domain of attraction of the Gumbel.
The norming constants can be taken $$ b_n\equiv F^{-1}(1-\frac{1}{n}) $$ and $$ a_n\equiv A(b_n) $$ where $A:\mathbb{R}\rightarrow (0,\infty)$ is called auxiliary function of $F$ and $F^{-1}$ denotes quantile function.
Auxiliary functions are not unique. If the pdf $f$ of $F$ exists, an auxiliary function is $$ A(x)\equiv \frac{1-F(x)}{f(x)} $$
Question: suppose that $F$ is Gumbel; what is $A$? I found in some sources that $A(x)=1$ but I don't understand how to show it. I am also confused because when I compute $$ A(x)\equiv \frac{1-F(x)}{f(x)}= \frac{1-e^{-e^{-x}}}{e^{-(x+e^{-x})}} $$ I don't get $1$.