Can the definition of regular languages be simplified?

Wikipedia says

The collection of regular languages over an alphabet Σ is defined recursively as follows:

• The empty language Ø is a regular language.
• For each a ∈ Σ (a belongs to Σ), the singleton language {a} is a regular language.
• If A and B are regular languages, then A ∪ B (union), A • B (concatenation), and A* (Kleene star) are regular languages.
• No other languages over Σ are regular.

I think that if it is closed under concatenation, then it must be closed under Kleene star, because we can take B to be A. So the definition can be without being closed under Kleene star?

• If you do not allow the star operation, you cannot obtain the language containing only the empty word. This language is equal to $\emptyset^*$. – J.-E. Pin Jul 22 '14 at 8:06

No. Closure under concatenation means that, if $A$ and $B$ are regular languages, then so is the language of strings formed by concatenating taking one string from $A$ and one from $B$. Closure under Kleene star means that, if $A$ is a regular language, then so is the langauge formed by taking any finite number of strings from $A$ and concatenating them.
If you delete Kleene star then every regular language would be finite. The reason is that any given regular language must be formed by a finite number of operations from the basic languages ($\emptyset$ and singletons). But any language whose definition involves no Kleene stars and only $k$ concatenations can only contain strings of length at most $k$, so it must be finite (it has at most $|\Sigma|^k$ strings in it).