To prove that

$\sum_{i=0}^{n} (-1)^n \binom{n}{i} = 0$

Step 1: Understand the summation The sum involves the binomial coefficient $\binom{n}{i}$, which represents the number of ways to choose $i$ elements from $n$, multiplied by the alternating sign factor $(-1)^n$. Let’s analyze it.

Step 2: Factor out $(-1)^n$ Since $(-1)^n$ is independent of $i$, we can factor it out of the summation: $\sum_{i=0}^{n} (-1)^n \binom{n}{i} = (-1)^n \sum_{i=0}^{n} \binom{n}{i}.$

Step 3: Evaluate $\sum_{i=0}^{n} \binom{n}{i}$ From the binomial theorem, we know that: $\sum_{i=0}^{n} \binom{n}{i} = (1 + 1)^n = 2^n.$

Thus: $\sum_{i=0}^{n} \binom{n}{i} = 2^n.$

Step 4: Substitute back Now substitute this result into the original equation: $\sum_{i=0}^{n} (-1)^n \binom{n}{i} = (-1)^n \cdot 2^n.$

Step 5: Evaluate $(-1)^n \cdot 2^n$ The expression $(-1)^n \cdot 2^n$ is simply $-2^n$ when $n$ is odd, and $+2^n$ when $n$ is even.

Correction to the question:

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