What is the meaning of ∀x∃x? Say we are in first-order logic, $x$ is a variable and its values are from a set of values, doesn't matter what.
$∀x∃x$ means for every $x$ (from the set of values) there exist at least one x (from the set of values). Since we took all $x$s from the set, of course each one itself does exist for itself because $x$ is $x$. Hence this should actually shorten to $∀x$ am I right?
The same reasoning for $∃x∀x$: there exist at least one $x*$ such that every $x$ from those $*$ is $x$. Well of course, since $x$ is $x$. So it shortens to $∃x$. Am I correct?
Conclusion
Thank you for all the answers. I was wrong, it is the other way around. I think the gist of it is that the innermost quantifier matters only when quantifying the same variable for the same formula, and the outer quantifiers (of the same variable & formula) become null quantifiers since there is nothing to quantify anymore. This is, analog to computer programs, similar to variable shadowing, which is when you over-declare/define a variable, here we are over-defining quantifiers in an expression like the above.
 A: You are right that with multiple quantifiers that are next to each other and that quantify the same variable, you can reduce it to just one quantifier, as the others effectively end up not doing any interesting work at all; they are what are called 'null quantifiers'.
However, you got it just the other way around. As it turns out, the quantifier that stays is the most inside quantifier, and the ones outside of it become the null quantifiers.
That is, a statement like $\forall x \ \exists x \ P(x)$ is the same as $\exists x \ P(x)$
And $\exists x \ \forall x \ P(x)$ is equivalent to $\forall x \ P(x)$
To explain: 
Take $\forall x \exists x \ P(x)$ . Now, suppose that there is at least one object $x$ such that it has property $P$. Then $\exists x \ P(x)$ is true, and notice that that statement has no free variables $x$. So, when you 'quantify' that expression by putting a $\forall x$ in front of it, then that quantifier does not quantify anything and is therefore a null quantifier, and can be removed without changing the truth-conditions of the sentence
A: A more succinct explanation: it's obvious that $\exists x P(x)$ is equivalent to $\exists y P(y)$. You can't change the truth value of a statement merely by renaming some entities.
So $\forall x [\exists x P(x)]$ should be equivalent to $\forall x [\exists y P(y)]$ - if you're queasy about this, step through it explicitly by setting $\phi := \exists x P(x)$.
But that is manifestly equivalent to $\exists y P(y)$ (assuming as usual that the domain of discourse is not empty), which is just $\exists x P(x)$.
