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Factorial notation is a mathematical concept used to denote the product of an integer and all the non-negative integers below it. It is represented by an exclamation mark (!). For example, the factorial of 5, written as 5!, is calculated as:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Understanding how factorial notation works is crucial when dealing with factorial equations. In such equations, expanding factorials by their definition often simplifies the problem at hand, much like in the exercise where the factorial of \((n + 1)\) is expanded to \((n + 1) \cdot n!\). Remember, for any positive integer \(n\), \(n!\) is the multiplication of all positive integers from \(1\) up to \(n\). Thus, \((n + 1)!\) includes all the terms in \(n!\) with an extra multiplication by \((n + 1)\).

Algebraic manipulation refers to the process of rearranging and simplifying algebraic expressions and equations to resolve them or present them in a more understandable form. This usually involves a combination of various techniques such as factoring, distributing, combining like terms, and applying inverse operations.
In our exercise, algebraic manipulation comes into play when the factorial expression \((n + 1)!\) is broken down. Realizing that \((n + 1)!\) can be expressed as \((n + 1) \cdot n!\) is an instance of factoring. By recognizing factorial properties, we're able to simplify the ratio \(\frac{(n+1) !}{n !}\) to \((n + 1)\), which is a much simpler algebraic expression to work with. Mastering such manipulation tactics is key for efficiently solving equations with factorials.

Solving equations is the process of finding the unknown variables that make the equation true. In the context of our provided exercise, solving the equation begins once we have simplified the factorial components and are left with the basic algebraic equation \(n + 1 = 10\).
To solve for \(n\), one needs to perform inverse operations that will isolate \(n\) on one side of the equation. This involves subtracting 1 from both sides to undo the addition of 1 to \(n\). This gives us:
\[ n = 10 - 1 \]\[ n = 9 \]
This step—performing an inverse operation—is a foundational concept in algebra. Equations can often be resolved by understanding the 'undoing' process, whether it be subtraction to undo addition, division to undo multiplication, or more complex operations necessary for higher-level algebra.