Show that $T \mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^{2n}$

Question

$T \mathbb{R}^n$ is the tangent bundle to $\mathbb{R}^n$. How do I show that is is diffeomorphic to $\mathbb{R}^{2n}$? I know that there is an isomorphism (and therefore a diffeomorphism) between $\mathbb{R}^n$ and $T_p \mathbb{R}^n$, can I conclude from here, using the fact that $TM$ is the disjoint union of all $T_pM$s?

Alternatively: I know that, if $M$ is an $n$-dimensional differentiable manifold, then $TM$ is a $2n$-dimensional differentiable manifold. I have seen the construction given in Warner, Foundations of Differentiable Manifolds and Lie Groups, page 19, complete with proof. It really shouldn't be hard to explicitely construct the diffeomorphism, but I'm struggling for ideas.

Answer 1

If $M$ is an $n$-dimensional manifold defined by the atlas $(U_i,f_i:U_i\rightarrow\mathbb{R}^n)_{i\in I}$ then $TM$ is the manifold defined by $(U_i\times\mathbb{R}^n,g_i)$ where $g_i(x,y)=f_i\circ f_j^{-1}(x), d(f_i\circ f_j^{-1})_{f_j(x)}(y))$. $\mathbb{R}^n$ is an $n$-dimensional manifold defined by the chart $(\mathbb{R}^n, Id_{\mathbb{R}^n})$, therefore $T\mathbb{R}^n$ is defined by the chart $(\mathbb{R}^n\times\mathbb{R}^n, Id_{\mathbb{R}^{2n}})$.

Answer 2

Going by Lee's definition, the tangent bundle is, by definition, disjoint union of all $T_pM $s.
Let $(U,\phi)$ be in atlas of M. Define map $f: \pi^{-1}(U) \rightarrow R^{2n} $ as $$f\Bigg(v^i|\frac{\partial}{\partial x^i}\Bigg|_p \Bigg)=(p,v) $$ Now write down the inverse of $f$ and observe that it is a bijection. Now write down the map $ id\circ f\circ\phi^{-1}$. This should in fact turn out to be identity map.