Surface integral vs contraction

Question

I have to prove that there exists a contraction $f:\mathbb R^{n-1}\to S^{n-1}$ with $S^{n-1}=\partial B_1(0)\subset \mathbb R^n$. Now, it is well known that the surface of a radius one $\mathbb R^n$ ball has finite "area" wich I think can be written as $$A = \frac{2\,\pi^\frac{n+1}{2}}{\Gamma\left(\frac{n+1}{2}\right)} $$ is it true that since $A$ is finite we can define a contraction between $\mathbb R^{n-1}$ and $S^{n-1}$? I think that a positive answer should be based on the surface integral which computes $A$, but I don't know where to start.