Proving that the volume of a unit ball is bounded by $c (\frac{c'}{d})^{\frac{d}{2}}$

Question

Given that $B^d \subseteq \sqrt{d}B_1^d$, how do I use this to prove that the unit ball is bounded by $c (\frac{c'}{d})^{\frac{d}{2}}$ where $c, c'$ is some constant and $c, c' > 0$. Here $B^d$ is the unit ball and $B_1^d$ is a cross-polytope. The volume of $\sqrt{d}B_1^d = d^{\frac{d}{2}}2^d(d!)^{-1}$. What is the best way to approach this problem?

Answer 1

Stirling's approximation implies $$0<\sqrt{2\pi} d^{d+1/2}e^{-d} \leq d!~~,$$ and we have \begin{eqnarray} vol(B^d)&\leq&vol(\sqrt{d}B_1^d)\\ &=&\frac{d^{\frac{d}{2}}2^d}{d!}\\ &\leq&\frac{d^{\frac{d}{2}}2^d}{\sqrt{2\pi} d^{d+1/2}e^{-d}}\\ &=&\frac{2^de^d}{\sqrt{2\pi} d^{d/2+1/2}}\\ &\leq&\frac{2^de^d}{\sqrt{2\pi} d^{d/2}}\\ &=&\frac{(4e^2)^{d/2}}{\sqrt{2\pi} d^{d/2}}\\ &=&\frac{1}{\sqrt{2\pi}}\cdot\Big(\frac{4e^2}{d}\Big)^{d/2}, \end{eqnarray}

$\therefore~c=\frac{1}{\sqrt{2\pi}}$ and $c'=4e^2$.