The equation to be proved is: $\sum_{j=0}^n(-1)^j\dbinom{n}{j} = \dbinom{n}{0} - \dbinom{n}{1} + ... \pm \dbinom{n}{n}=0$
But if i take the base case ($n = 1$) i get
$\sum_{j=0}^n(-1)^j\dbinom{n}{j} = \dbinom{n}{0} - \dbinom{n}{1}$
But this evaluates to $\frac{n!}{0!n!} - \frac{n!}{1!(n-1)!}$
I don't think the above statement evaluates to zero. Could this be an error? The book is Spivak's Calculus and the problem is 2-3(e)(ii)
Recall the binomial formula: $$(1+x)^n = \sum_{j=0}^n \dbinom{n}j x^j$$ Taking $x=-1$, we obtain what we want.
There is also a bijective proof. Recall that $\binom{n}{k}$ counts the number of subsets of $[n] = \{1, ..., n\}$ with $k$ elements. Define $\text{sgn} : 2^{[n]} \to \{-1, 1\}$ by $\text{sgn}(S) = (-1)^{|S|}$. That is, $\text{sgn}(S) = 1$ if $|S|$ is even and $\text{sgn}(S) = -1$ if $|S|$ is odd. Observe that exactly half of the elements of $2^{[n]}$ contain $1$. So define a bijection: $\phi : 2^{[n]} \to 2^{[n]}$ by $\phi(S) = S \triangle \{1\}$. Observe that $\text{sgn}(S) + \text{sgn}(\phi(S)) = 0$.
It follows that $\sum_{i=0}^{n} (-1)^{i} \binom{n}{i} = \sum_{S \in 2^{[n]}} \text{sgn}(S) = 0$.
Linear-Algebraic Proof:
Let $n$ be a positive integer and $V:=\mathbb{F}_2^n$. For a vector $v:=\left(v_1,v_2,\ldots,v_n\right)\in V$, define $\phi(v):=v_1+v_2+\ldots+v_n$. Then, $\phi:V\to\mathbb{F}_2$ is a linear map. Clearly, $\phi$ is surjective so we have by the Rank-Nullity Theorem that $\dim_{\mathbb{F}_2}\big(\ker(\phi)\big)=\dim_{\mathbb{F}_2} (V)-\dim_{\mathbb{F}_2} \big(\text{im}(\phi)\big)=n-1$. That is, $\big|\ker(\phi)\big|=2^{n-1}$. As $|V|=2^n$, there are exactly $2^n-2^{n-1}=2^{n-1}$ elements of $V$ with the images under $\phi$ being $1$. Consequently, $$\left|\phi^{-1}(0)\right|=\big|\ker(\phi)\big|=2^{n-1}=\left|\phi^{-1}(1)\right|\,.$$ As $\left|\phi^{-1}(0)\right|=\sum_{r=0}^n\,\binom{n}{2r}$ and $\left|\phi^{-1}(1)\right|=\sum_{r=0}^n\,\binom{n}{2r+1}$, we have the required equality.
