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Understanding how to \textbf{expand factorials} is crucial when dealing with factorial expressions. Factorials, denoted by the exclamation mark (!), are products of all positive integers up to a given number. For instance, the factorial of 3, expressed as 3!, is calculated as 3 × 2 × 1, which equals 6.

When expanding factorial expressions like \( (n + 2)! \), you start from the given number and multiply it by each previous positive integer until you reach 1. However, you can also stop the multiplication once you reach a factorial term present in the denominator, because this anticipates the next concept of simplification through canceling common terms.

\textbf{In practice:}\
Let's expand the expression \( (n + 2)! \). It translates to:\
\[ (n + 2)(n + 1)n(n - 1)(n - 2)...(3)(2)(1) \].\
However, if \( n! \) is present in the denominator, we only need to expand until the term \( n! \), as the subsequent multiplication would be redundant for simplification purposes.

The process of \textbf{algebraic simplification} of expressions involves reducing them to their simplest form. This is done by performing operations and using properties such as the distributive, associative, and commutative properties of multiplication and addition.

In the context of factorial expressions, algebraic simplification frequently leverages the property that factorials themselves are products of sequences of consecutive integers, which allow them to simplify directly when divided by each other, provided one factorial is a subset of the other.

For our expression \( \frac{(n + 2)!}{n!} \), after expanding and realizing that \( n! \) is contained within \( (n + 2)! \) makes the simplification straightforward:\
\[ \frac{(n+2)*(n+1)*n!}{n!} \] simplifies to \( (n + 2)(n + 1) \), because the \( n! \)'s cancel each other out.

A key step in simplification is \textbf{canceling common terms}, which involves dividing both numerator and denominator by the same number or expression to simplify the fraction. In the example of factorial expressions, this often means identifying and canceling whole factorial terms that appear in both the numerator and the denominator.

As seen in the exercise, after factorial expansion and identifying common terms, we can cancel \( n! \) in both the numerator and the denominator because every number in \( n! \), by definition, is a factor of \( (n + 2)! \). What's left is simply the additional terms from the expanded numerator that were not part of the \( n! \) factorial.

The simplification process reduces the expression significantly and eliminates the redundancy of the factorial term, leaving an easier, more manageable expression:\
\[ (n+2) * (n+1) \], which is the product of two consecutive integers.