An application of the multiplicativity of degree extensions 
Let $\mathbb{F}$ be a field and $\mathbb{F\not=R}$. I want to show that there does not exist a field extension $\mathbb{F}/\mathbb{R}$ of odd degree. Here $\mathbb{R}$ is the field of real numbers.

My attempt: pick $\alpha\in\mathbb{F}, \alpha\notin\mathbb{R}$. By multiplicative nature of degree extensions,  [$\mathbb{F:R}$]=[$\mathbb{F:R(\alpha)}$][$\mathbb{R(\alpha):R}$]. I have been trying to show that [$\mathbb{R(\alpha):R}$] is even to no avail.
If I try to show this by contradiction, then one of [$\mathbb{F:R(\alpha)}$] or [$\mathbb{R(\alpha):R}$] should be odd. But I again I get stuck.
 A: Suppose $F$ has odd degree over $\mathbb{R}$; let $\alpha\in F$. Then, as you note,
$$[F:\mathbb{R}] = [F:\mathbb{R}(\alpha)][\mathbb{R}(\alpha):\mathbb{R}].$$
Since $[F:\mathbb{R}]$ is odd, then so is its divisor $[\mathbb{R}(\alpha):\mathbb{R}]$. Therefore, the minimal irreducible polynomial of $\alpha$ over $\mathbb{R}$ has odd degree.
If $p(x) = x^{2n+1}+\cdots + a_1x + a_0$ is the minimal irreducible polynomial for $\alpha$, then note that $\lim\limits_{x\to\infty}p(x) =\infty$ and $\lim\limits_{x\to-\infty}p(x)=-\infty$. By the intermediate value theorem, $p(x)$ has a root $c$. Therefore, by the factor theorem $x-c$ divides $p(x)$. Since $p(x)$ is irreducible and monic, $p(x)=c-x$, so $\alpha=c$. Therefore, $\alpha\in\mathbb{R}$.
Since $\alpha\in\mathbb{R}$ for all $\alpha\in F$, we conclude that $F\subseteq\mathbb{R}$. Hence, $F=\mathbb{R}$. So the only extension of $\mathbb{R}$ of odd degree is $\mathbb{R}$ itself, which has degree $1$.
A: Here is an application of that formula to a problem similar to yours above:


There is no extension of $\Bbb{R}$ of degree 4.


Suppose we have an extension $F$ of the reals of degree 4. Then  such an extension $F$ is $\Bbb{R}[a_1,\ldots a_n]$ where the $a_i \in F$ (why?).  Now since $F$ is an extension of $\mathbb{R}$ of degree 4 we are adjoining at most two elements of $F$ to $\Bbb{R}$. This is because if say we adjoin three elements $a,b,c$ to $\Bbb{R}$, without loss of generality we have
$$[\Bbb{R}(a,b,c) : \Bbb{R}] = [\Bbb{R}(a,b,c) : \Bbb{R}(a,b)   ] [\Bbb{R}(a,b), \Bbb{R}(a) ] [\Bbb{R}(a):\Bbb{R}] $$
or that we have 3 positive integers multiplying together to give 4. Therefore one of those extensions must have been the trivial extension. Similarly adjoining 4 or more elements does not create any bigger field than adjoining two elements. We now show that assuming $F$ is an extension of $\Bbb{R}$ by adjoining one or two elements of $F$ leads to a contradiction. Suppose $F = \Bbb{R}(a)$. Then this means that $a \in F$  is the root of a real monic irreducible polynomial of degree 4. However this is not possible because every real polynomial of degree 4 is reducible over $\Bbb{R}$ (exercise; use the fundamental theorem of algebra). So the only possibility left is that $F$ is obtained from $\Bbb{R}$ by adjoining two elements, which we call $\alpha $ and $\beta$. Write
$$[\Bbb{R}(\alpha,\beta) : \Bbb{R}] = [\Bbb{R}(\alpha,\beta):\Bbb{R}(\alpha)][\Bbb{R}(\alpha):\Bbb{R}].$$ 
For the right hand side to equal 4 this means that each factor in the right hand side is 2. We cannot have one of them being $1$ because then the other is 4 contradicting the fact that all real polynomials are reducible over $\Bbb{R}$. Now if $[\Bbb{R}(\alpha):\Bbb{R}]$ is 2 then this means that $\alpha$ is a root of a real polynomial of degree 2, so that $\alpha$ is contained in the complex numbers, hence $\Bbb{R}(\alpha) \subset \Bbb{C}$. But then we know  that there are no proper fields of finite degree as an extension of $\Bbb{R}$ between $\Bbb{C}$ and $\Bbb{R}$. This forces $\Bbb{R}(\alpha) = \Bbb{C}$. 
It follows that the extension $\Bbb{R}(\alpha,\beta)/\Bbb{R}(\alpha)$ is an extension of $\Bbb{C}$ of degree two. However this means that $\beta$ is the root of an irreducible polynomial of degree 2 with complex coefficients. However since  $\Bbb{C}$ is algebraically closed the only irreducible polynomials in $\Bbb{C}$ are those of degree 1. This is a contradiction so that the right hand side of equation $(1)$ can never equal 4. It follows that there is no field extension of $\Bbb{R}$ of degree 4. 
