This question is regarding the dimension of the tangent space $T_p\mathbb{R}^n$ as it is defined in the context of smooth manifolds. One the one hand, $T_p \mathbb{R}^n$ can just be interpreted as a copy of $\mathbb{R^n}$ based at a chosen point $p \in \mathbb{R}^n$. Now, since this is just $\mathbb{R}^n$ translated through $p$, clearly $T_p \mathbb{R}^n$ has the same dimension as $\mathbb{R}^n$. To see this, just take the canonical basis $\{e_1, \dots, e_n \}$ translate it through $p$ and note that this will yield a basis of $T_p \mathbb{R}^n$. Thus, $dim( T_p \mathbb{R}^n) = n$
On the other hand though, for a given $p$, it is common to define the tangent space of $\mathbb{R}^n$ at $p$ as $T_p\mathbb{R}^n = \{(p,x) | x \in \mathbb{R}^n \}$ Here, $T_p\mathbb{R}^n$ is effectively a Cartesian product of the one point vector space $\{p\}$ and $\mathbb{R}^n$. By this reasoning,
$ \displaystyle dim(T_p\mathbb{R}^n) = dim(\{p\} \times T_p \mathbb{R}^n) = dim(\{p\}) + dim(\mathbb{R}^n) = 1 + n $
Obviously, this last equation can't be correct but I'm not really sure where my reasoning is off. So, specifically, my question is where is the error in my logic?