How is this way of rewriting extreme-value problems a simplification?

This question is about pp. 370-374 of Harald Cramer's 1946 Mathematical Methods of Statistics. The author considers a more general question, but for simplicity let us focus on the question of:

Given i.i.d. observations $$X_1, \dots, X_n$$, what is the distribution of $$\max_{1 \le i \le n} X_i$$?

Question: The author considers the following approach. How does it in anyway present a simplification of the problem? Isn't the random variable $$\Xi$$ much more complicated than $$\max_{1 \le i \le n}X_i$$? For example, how does the author calculate its expectation when $$X_i \sim \operatorname{Unif}(a,b)$$?

Author's Approach: If the CDF of each $$X_i$$ is $$F(x)$$, then the CDF of $$\max_{1 \le i \le n}X_i$$ is $$F(x)^n$$, and the author assumes the $$X_i$$ are absolutely continuous with respect to Lebesgue measure, so the $$X_i$$ have a density function $$f(x)$$ with $$F'(x) = f(x)$$. Therefore, $$\max_{1 \le i \le n} X_i$$ has the density $$nF(x)^{n-1}f(x)$$ according to the chain rule (28.6.1).

Now introduce a new random variable

$$\Xi := n \left( 1 - F \left( \max_{1 \le i \le n} X_i \right) \right) \,. \tag{28.6.2}$$

Via a change of variables (see this CV question), the author concludes that the density of $$\Xi$$ is:

$$h(\xi) := \left( \frac{\xi}{n} \right)^{n-1} I_{[0,n]}(\xi) \,, \tag{28.6.3}$$ where $$I_{[0,n]}$$ denotes the indicator function of the interval $$[0,n]$$.

The author then proposes finding exact or asymptotic solutions for the problem of solving $$\max_{1 \le i \le n} X_i$$ in terms of $$\Xi$$, with the belief that this amounts to a simplification of the problem.

Note: This is an indirect follow-up to a previous question of mine.

Note: In the course of writing up this question, I realized the answer. In particular, what the author was saying went over my head because the notation was unusual to me. In the course of changing the notation to make it readable to anyone who might answer the question, I understood finally what the author was saying. In order to make this not a complete waste of time, I am typing up my realization as a community wiki answer so others can avoiding wasting their time the same way.

Basic Idea: This is sort of a "extreme value version of inverse transform sampling". In particular, for any continuous distribution function $$F$$, the random variable $$\Xi$$ has the exact same distribution.

Details: In particular, the distribution of $$\Xi$$ can be seen to be related to the distribution of the maximum of $$n$$ i.i.d. uniform random variables. More importantly the distribution of $$\Xi$$ has a simple closed-form density function and therefore moments and other quantities of $$\Xi$$ are simple to calculate. Thus writing $$\max_{1 \le i \le n}X_i$$ in terms of $$\Xi$$ makes at least the moments of $$\max_{1 \le i \le n}X_i$$ possibly easier to calculate, and other quantities as well in some cases.

Regarding the $$\operatorname{Unif}(a,b)$$ example I mentioned earlier, the author shows that

$$\max_{1 \le i \le n }X_i = b - \frac{b-a}{n} \Xi \,.$$

The claim I didn't understand was why this implied that:

$$\mathbb{E} \left[ \max_{1 \le i \le n } X_i \right] = b - \frac{n}{n+1}(b-a)\,,$$

i.e. why one has that

$$\mathbb{E} [ \Xi] = \frac{n^2}{n+1} \,.$$

However, it is easy to see with basic calculus that:

$$\mathbb{E} [ \Xi] = \frac{1}{n^{n-1}} \int_0^n (\xi)(\xi^{n-1}) d \xi = \frac{n^2}{n+1} \,.$$

It is obvious to me now as written, but the notation used in the book was originally unclear to me.