Show that $\lambda^n(B_n^p) \le C(p)$ 
Let $p \in (0, \infty)$, $n \in \Bbb N$, $\lambda$ the Lebesgue-measure and $B_n^p := \{x \in \Bbb R^n : ||x||_p \le 1\}$.
Show that there exists a constant $C(p) \in (0, \infty)$ that only depends on $p$, such that for every $n \ge n_0(p)$, the inequlity
$\lambda^n(B_n^p) \le C(p)$.
Hint: Use mathematical induction.

I'm not sure what I am expected to do here. I found out that $\lambda^n(B_n^p)$ can be written as
$1 \over k! \pi^k$, when $k = n$ is even, and ${2^{k+1} \over 1 * 3 * ... * (2k + 1)} \pi^k$, when $k = n$ is odd, but that's about it so far. I don't know how to work with the $p$ in this scenario. And aren't all $p$-norms equivalent anyway?
 A: Compute $\lambda^n(B_n^p)$ in terms of the $\lambda^{n-1}$ measure of its slices:
$$\lambda^n(B_n^p)=\int_{-1}^1 \lambda^{n-1}(B_n^p\cap \{x_n=t\})dt$$
where $x_n$ is the last coordinate in $\mathbb R^n$. We have that for fixed $t$
$$B_n^p\cap \{x_n=t\}=\{(x_1,\ldots,x_{n-1},t):|x_1|^p+\ldots+|x_{n-1}|^p+|t|^p=1\}$$
is equivalent to a ball in $(\mathbb R^{n-1},\|\cdot\|_{p})$ of radius $(1-|t|^p)^{1/p}$. Using the homogeneity of the Lebesgue measure then
$$\lambda^{n-1}(B_n^p\cap\{x_n=t\})=(1-|t|^p)^{\frac{n-1}{p}}\lambda^{n-1}(B_{n-1}^p).$$
Therefore
$$\lambda^n(B_n^p)=\lambda^{n-1}(B_{n-1}^p)\int_{-1}^1 (1-|t|^p)^{\frac{n-1}{p}}dt$$
and if we can show that for $n$ large enough the last integral is $\leq 1$ we are done. But
$$f_n(t)=(1-|t|^p)^{\frac{n-1}{p}}$$
is a sequence of bounded functions that converge pointwise to $0$ in $[-1,1]\backslash\{0\}$ as $n\to\infty$, and therefore their integrals will converge to $0$ by dominated convergence, and thus will be eventually smaller than $1$.
If $p=\infty$ this proof breaks down and in fact $\lambda^n(B_n^\infty)=2^n$.
