Proof By Induction Involving Multiple Inequalities Let $m,n \in\mathbb{N}$, with $m<n$. Show that
$$\frac{1}{m^k} \binom{m}{k}<\frac{1}{n^k}\binom{n}{k} \leq \frac{1}{k!}\leq \frac{1}{2^{k-1}}$$ for all ${2 \le k \le m}$.
I know to prove this by induction and have established the base case for $k = 2$, and have assumed true for $k = \alpha$ where $\alpha \ge 2$. However, I do not understand how to structure the inductive step (let $k = \alpha + 1$) given that there are three inequalities and four statements. I would be very appreciative of some help on how to prove statements inductively with multiple inequalities.
 A: The inequality $\displaystyle\frac{1}{m^k}\binom mk<\frac{1}{n^k}\binom nk$ is trivial for $n>m$ and no induction is required. For the second, we have for $2\le k\le m:$
$$\frac{k!}{n^k}\binom{n}{k}=\frac{n!}{n^k(n-k)!}=\frac{\prod_{i=1}^k(n-i+1)}{n^k}\le1$$
Because each term in the product is $\le n$. This yields $\displaystyle\frac{1}{n^k}\binom nk<\frac{1}{k!}$, again with no induction. Finally, we can procede by induction for the last inequality. For the base case $k=2$, we notice that the inequality is actually not strict, but we can show that it is for $k\ge3$. For $k=3$, the strict inequality holds. Now suppose that it is the case for some integer $k\ge2$. Therefore, we have :
$$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}<\frac{1}{2^{k-1}}\cdot\frac1{k+1}<\frac{1}{2^{k-1}}\cdot\frac12=\frac{1}{2^k}$$
Which concludes the induction.
By combining all the obtained inequalities, we therefore get for $2\le k\le m$:
$$\displaystyle\frac{1}{m^k}\binom mk<\frac{1}{n^k}\binom nk\le\frac{1}{k!}\le\frac{1}{2^{k-1}}$$
And the last inequality is strict if $k\ge3$
