Why does $\lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n$ if $ \Bbb{Q} [\sqrt[n]{2}]$ is formed by adding only one element? I know that the degree of an extension the dimension of, in this case, $\Bbb{Q} [\sqrt[n]{2}]$ seen as a vector space over  $\Bbb{Q} $. But I thought that the degree of an extension field formed by adding just one element to a given field was one. So, I find it confusing that 

$\lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n$

Given that $ \Bbb{Q} [\sqrt[n]{2}]$ is formed by adding only one element, namely $\sqrt[n]{2}$, to the field of rational numbers. Any help would be very much appreciated. Thank you.
You are confusing the vector space obtained by adding one element with the field obtained by adding only one element.

Since $\mathbb Q[\sqrt[n]{2}]$ is a field, when we add $\sqrt[n]{2}$ we are also adding  $\sqrt[n]{2}\cdot  \sqrt[n]{2},...,\sqrt[n]{2}\cdot...\cdot\sqrt[n]{2}$ where in the last term there are $n-1$ products. It is enough to stop here since the product of n terms 
$$\sqrt[n]{2}\cdot...\cdot\sqrt[n]{2}=2 \in \mathbb Q$$
is already here.
$\mathbb{Q}[\sqrt[n]{2}]$ is a vector space over the field $\mathbb{Q}$. The equation
$$[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}]=n$$
says that the dimesnion of this vector space over $\mathbb{Q}$ is $n$. 

In general, if $K$ and $L$ are fields such that $K\subseteq L$, then $L$ can be considered as vector space over $K$. Then $[L:K]$ denotes the dimension of this vector space.