Show that $p(n, m)=\frac{m-1}{m} \times \cdots \times \frac{m-n+1}{m} \le (e^{\frac{-n}{2m}})^{n-1}$ Suppose that $n,m \in \mathbb{Z}$. Show that $p(n, m) \le e^{-\frac{n(n-1)}{2m}}$. It's also given that $p(n,m)= \frac{m-1}{m} \times \cdots \times \frac{m-n+1}{m}$.
It's also given that $n$ is odd, so $(n-1)/2 \in \mathbb{N}$.
So we will try to find an upper bound:
$$(m-1) \times \cdots \times ( m-n+1) = \prod_{k=1}^{n-1}(m-k)$$
This can be bounded as,
\begin{align}
\prod_{k=1}^{n-1}(m-k) &= \prod_{j=1}^{\frac{n-1}{2}}(m-j) \times (m-n+j) \tag{1}\label{tag1}\\
&\vdots\\
&\le \prod_{j=1}^{\frac{n-1}{2}}\left(m-\frac{n}{2}\right)^2 \\&= \left(\left(m-\frac{n}{2}\right)^2\right)^{\frac{n-1}{2}} \\&= \left(m-\frac{n}{2}\right)^{n-1} 
\end{align}
We can conclude that,
$$p(n,m) \le \left(\frac{m-\frac{n}{2}}{m}\right)^{n-1} = \left(1-\frac{n}{2m}\right)^{n-1} $$
$$\left(1-\frac{n}{2m}\right)^{n-1}  \le (e^{\frac{-n}{2m}})^{n-1} \tag{2}\label{tag2}\\$$
Problems:

*

*Can you please show how \ref{tag1} changed bound of product to $\frac{n-1}{2}$ and how we got terms inside $(m-j)(m-n+j)$ please?


*Can you please show how \ref{tag2} got upper bound $e^{\frac{-n}{2m}} $ for $\left(1-\frac{n}{2m}\right)$?
 A: Your proof is mostly fine.
There is a problem that your proof assumes $p(m,n)$ has an even number of terms. The combining of terms in $(1)$ is only true when $n$ is odd.
It is easier to prove, when using this method, that $$p^2(m,n)\leq e^{n(n-1)/m}$$
Then $$\begin{align} 
m^{2(n-1)}p^2(m,n)&=\left(\prod_{k=1}^{n-1}(m-k)\right)^2\\
&= \left(\prod_{k=1}^{n-1}(m-k)\right) \left(\prod_{k=1}^{n-1}(m-n+k)\right)
\\&=\prod_{k=1}^{n-1}(m-k)(m-n+k)
\end{align} 
$$
and then use the AM/GM on these terms, as in your posted proof. The second equality is because $$\prod_{k=1}^{n-1} f(k)=\prod_{k=1}^{n-1}f(n-k)$$
The last step, $(2)$ in your proof, is because $0<1+x\leq e^x,$ where $x=-\frac{n}{2m}.$

An easier proof uses Bernoulli’s inequality: $$1+kx\leq (1+x)^k, x\geq -1, k\in \mathbb N$$
and the related inequality:
$$1+x\leq e^x, x\in \mathbb R.\tag3$$
We assume $n\leq m.$ If $n>m,$ the inequality is true because $p(m,n)=0.$
For $1\leq k<m,$ Bernoulli shows, for $x=-\frac 1m$:
$$0<1-\frac km\leq \left(1-\frac1m\right)^k$$
So:
$$\begin{align} 
p(m,n)&=\left(1-\frac1m\right) \left(1-\frac2m\right)\cdots \left(1-\frac{n-1}m\right)\\
&\leq \left(1-\frac1m\right)^{1+2+\cdots+(n-1)}\\
&= \left(1-\frac1m\right)^{n(n-1)/2}
\end{align}$$
Finally, by $(3)$ for $x=-\frac1m,$ we have:
$$1-\frac1m\leq e^{-1/m}.$$
So: $$p(n,m)\leq e^{\frac{-n(n-1)}{2m}}$$

Bernoulli can be used to prove $(3)$ since Bernoulli shows that: $$1+x\leq \left(1+\frac xk\right)^k$$ for $k>|x|.$ So:$$1+x\leq \lim_{k\to\infty} \left(1+\frac xk\right)^k=e^x$$
