Prove that a multilinear function $f$ is skew-symmetric if and only if $f = \mathrm{Alt}(f)$ 
Prove that a multilinear function $f$ is skew-symmetric if and only if $f = \mathrm{Alt}(f)$.

I said the first thing is to prove that $\mathrm{Alt}(f)$ is skew-symmetric. In other words, we want to see if $\mathrm{Alt}(f(v_{\sigma(1)}, \cdots v_{\sigma(q)})) = (-1)^{\sigma}\mathrm{Alt}(f(v_1 \cdots v_q))$. So to do this we say $\mathrm{Alt}(f(v_{\sigma(1)}, \cdots v_{\sigma(q)})) =$
$$\frac{1}{q!}(-1)^{\sigma} \sum_{\mu \in S_1} (-1)^{\mu}(-1)^{\sigma} f(v_{\sigma \mu(1)}, \cdots v_{\sigma\mu(q)}) = \frac{1}{q!}(-1)^{\sigma} \sum_{\mu \in S_q} (-1)^{\sigma \mu} f(v_{\sigma\mu(1)}, \cdots v_{\sigma\mu(q)})).$$
As $\mu$ runs through $S_q$, we can see that $\sigma \mu$ also runs through $S_q$. Letting $\tau = \sigma \mu$, we get
$$\frac{1}{q!}(-1)^{\tau} \sum_{\tau \in S_q} (-1)^{\tau} f(v_{\tau(1)}, \cdots v_{\tau(q)})) = \mathrm{Alt}(f(v_1 \cdots v_1)).$$
Now we want to show that $f$ is skew symmetric. So to do this we simply say
$$\mathrm{Alt}(f(v_1 \cdots v_1)) = \frac{1}{q!}\sum_{\sigma \in S_q} (-1)^{\sigma} f\underbrace{(v_{\sigma(1)}, \cdots v_{\sigma(q)})}) $$
so, as $f$ is skew symmetric we can say that the bit in the underbrace is equal to $(-1)^{\sigma}f(v_1, \cdots v_q)$, and so the RHS of that equality is equal to 
$$\frac{1}{q!}\sum_{\sigma \in S_q} f(v_1 , \cdots , v_q) = f(v_1, \cdots v_q)$$
which completes the proof.
Is this correct?
 A: The first thing to do is show that $\mathrm{Alt}(f)$ is skew-symmetric. In other words, we want to see if $\mathrm{Alt}(f(v_{\sigma(1)}, \cdots v_{\sigma(q)})) = (-1)^{\sigma}\mathrm{Alt}(f(v_1 \cdots v_q))$. To do this we say, by definition, $\mathrm{Alt}(f(v_{\sigma(1)}, \cdots v_{\sigma(q)})) =$
$$\frac{1}{q!}(-1)^{\sigma} \sum_{\mu \in S_1} (-1)^{\mu}(-1)^{\sigma} f(v_{\sigma \mu(1)}, \cdots v_{\sigma\mu(q)}) = \frac{1}{q!}(-1)^{\sigma} \sum_{\mu \in S_q} (-1)^{\sigma \mu} f(v_{\sigma\mu(1)}, \cdots v_{\sigma\mu(q)})).$$
As $\mu$ runs through $S_q$, we can see that $\sigma \mu$ also runs through $S_q$. Letting $\tau = \sigma \mu$, we get
$$\frac{1}{q!}(-1)^{\tau} \sum_{\tau \in S_q} (-1)^{\tau} f(v_{\tau(1)}, \cdots v_{\tau(q)})) = \mathrm{Alt}(f(v_1 \cdots v_q).$$
This shows that $\mathrm{Alt}(f)$ is skew-symmetric. Therefore $f = \mathrm{Alt}(f)$ implies that $f$ is skew symmetric. 
Now we want to see that if $f$ is skew symmetric, then $f = \mathrm{Alt}(f)$. To do this we simply say
$$\mathrm{Alt}(f(v_1 \cdots v_1)) = \frac{1}{q!}\sum_{\sigma \in S_q} (-1)^{\sigma} \underbrace{f(v_{\sigma(1)}, \cdots v_{\sigma(q)})} $$
so, as $f$ is skew symmetric we can say that the bit in the underbrace is equal to $(-1)^{\sigma}f(v_1, \cdots v_q)$, and so the RHS of that equality is equal to 
$$\frac{1}{q!}\sum_{\sigma \in S_q} f(v_1 , \cdots , v_q) = f(v_1, \cdots v_q)$$
which completes the proof by showing that $f = \mathrm{Alt}(f)$, which we know to be skew symmetric from the first bit.
