Step 1 :Let $a_{1}, a_{2}, \ldots, a_{n} \in \mathbb{R}$ be a sequence of real numbers.
Step 2 :Let $A=\{1,2,3, \ldots n\}, n \in \mathbb{Z}^{+}$, and $\pi: A \rightarrow A$ be an automorphism on $A$ (an automorphism is a bijection of a set onto itself).
Step 3 :Consider the sum of the sequence, $\sum_{i=1}^{n} a_{i}$. Since addition is a binary operation, we can only add two numbers at a time.
Step 4 :We claim that the sum of the sequence remains the same even if the order of the numbers in the sequence is changed. In other words, $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} a_{\pi(i)}$ for any permutation $\pi$ of the sequence $a_{1}, a_{2}, \ldots, a_{n}$.
Step 5 :This claim is a direct consequence of the commutative property of addition, which states that the order of numbers does not affect their sum.
Step 6 :\(\boxed{\text{Final Answer: } \sum_{i=1}^{n} a_{i} = \sum_{i=1}^{n} a_{\pi(i)}}\)