Diffeomorphism: Unit Ball vs. Euclidean Space In my differential geometry class we are being asked to prove that the open unit ball 
$B^n$ = { $x$ $\in$ $\mathbb{R}$$^n$ such that |$x$| < $1$}  
is diffeomorphic to $\mathbb{R}$$^n$
I am having a hard time with this as I am brand new not only to differential geometry, but also topology.
I know that I need to construct a smooth, differentiable bijection between the two with a differentiable inverse, but beyond that, I am unsure of where to start. Some guidance in the right direction would be greatly appreciated. 

Standard Approach
Consider the identification:
$$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_E^2}}$$
Its inverse is given explicitely by:
$$\Psi:\mathbb{R}^n\to\mathbb{B}^n:\Psi(y):=\frac{y}{\sqrt{1+\|y\|_E^2}}$$
The argument of the roots never vanish:
$$1-\|x\|_E^2\neq0\quad1+\|y\|_E^2\neq0$$
So they're both differentiable.
Alternative Approach
Consider the identification:
$$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{1-\|x\|_E^2}$$
By the inverse function theorem:
$$\mathcal{N}d\Phi(x)\equiv(0)\implies\mathrm{d}\Psi(y)=\mathrm{d}\Phi(\Psi(y))\in\mathcal{L}(\mathbb{R}^n)$$
But this holds globally as the identification is a bijection.
Problematic Approach
Consider the identification:
$$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_\infty^2}}$$
This is most likely not differentiable.
