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Understanding the power of a power property is crucial when dealing with more complex algebraic expressions that involve exponents. This rule is quite intuitive once you grasp the basic concept that when raising a power to another power, you simply multiply the exponents. Formally, the rule is expressed as \( (a^n)^m = a^{n*m} \) where \( a \) is the base and \( n \) and \( m \) are the exponents.

Let's see this in practice. If we have an expression like \( -\big(ab^3\big)^2 \), applying the power of a power property allows us to simplify the expression to \( -a^2b^6 \). We multiplied the exponents of \( b \) (3 and 2) to get the new exponent 6. It's important to keep in mind the base and its sign, as they can affect the final answer.

When you're faced with an algebraic expression that includes variables, such as \( a \) and \( b \) in our example, the substitution method is a technique that replaces those variables with specific numeric values. This step is often vital to evaluating expressions and comes after simplifying them algebraically as much as possible.

In our example, once we have simplified \( -\big(ab^3\big)^2 \) to \( -a^2b^6 \) using the power of a power property, we proceed to substitute \( a \) with 1 and \( b \) with 2. The process transforms the expression into \( -1^2 \times 2^6 \) which is ready for the final evaluation. It's essentially like filling in the blanks with given numbers to find out what the entire phrase (in this case, the expression) means.

Once we have simplified the expression and substituted in the given values, it's time for evaluating expressions. Evaluating means performing the actual computations to get a numerical result. This step requires careful attention to the order of operations, ensuring exponents are dealt with before multiplication, and keeping an eye on negative signs.

For the given exercise, after substitution, we evaluate \( -1^2 \times 2^6 \). Since \( 1^2 = 1 \), and \( 2^6 = 64 \) (because 2 multiplied by itself six times equals 64), the expression simplifies to \( -1 \times 64 \) which equals to \( -64 \). This is the final answer, signifying that the whole process from expressing powers of powers to substitution has led to a single numerical value that answers the question posed by the original algebraic expression.