While applying Nechiporuk’s Theorem to get a lower bound on the formula size of Element distinction function, the following inequality is used: (ref here)
$C(m^2,m-1) \geq (m^2-m+1)^{m-1}$.
Is there a general such inequality from which we are getting the above?
Question 2 : Also in the same proof I do not understand how to prove that if $n=2m\lceil \log m\rceil $ then $m \geq \frac{n}{\log n} $.
In fact the reverse inequality holds true almost all the time.
It's clear, as a related result, that $\binom{a+b}{a}\le (a+1)^b$ (with $a,b>0$) since the binomial coefficient calculation can be paired into $b$ terms: $$\begin{align}\binom{a+b}{b} &= \frac{(a+1)\cdot(a+2)\cdots(a+b)}{1\cdot 2\cdots b} \\ &=\frac{a+1}{1}\frac{a+2}{2}\cdots\frac{a+b}{b} \\ &= (a+1)(a/2+1)\cdots( a/b+1) \\ & \le (a+1)^b \end{align}$$
The result I want to show here is that $\binom{a+b}{b}\le a^b$ under most conditions. This is not always true, most obviously not when $b=1$, but since the multiplicative terms above are in decreasing order, it will be true for $b>1$ if $(a+1)(a/2+1)\le a^2$, implying $a^2/2 + 3a/2 + 1 \le a^2$ and thus $3a + 2\le a^2$, i.e. $a\ge 4$.
Referring back to the original formula with $a=m^2{-}m{+}1$ and $b=m{-}1$, requiring $b>1$ gives $m>2$ and in that case we always have $\binom{m^2}{m{-}1}\le(m^2{-}m{+}1)^{m{-}1}$ (and in fact the inequality is strict).