Approximation of multiplicity when Ideal gas low density is applied $\frac{M !}{(M-N)!} \approx M^{N}$ Our lecturer today mentioned how a piston's head being at equal pressure maximised the multiplicity of states. 

He mentioned the following:

If I have a fixed number of particles $N_A$ on left and $N_B$ on the right, and the whole system has a fixed total volume of $M$, we can say the movable piston partitions the total volume into $M_A$ and $M_B$ lattice sites respectively on each side, and therefore

$$M_{total} = M_{A}+M_{B} = constant$$
Hence if we want the volumes of the two sides that maximise the multiplicity function, 
$$\Omega(N, M)=\frac{M !}{(M-N) !} \frac{1}{N !}$$
This all made perfect sense. 
The next step is where I have confusion. 

As we are dealing with an ideal gas, the densities are so low that $\frac{N}{M} << 1$ hence

$$\frac{M !}{(M-N)!} \approx M^{N}$$
I understand the approximation as there are fewer molecules N than there are sites M for an ideal gas, but I fail to understand how he managed to reach the conclusion above from the previous equation above that. 
I tried to solve this using the approximation:
$$x! = \left(\frac{x}{e}\right)^x$$ but this failed to reach me to the correct working in order to prove the $\approx M^{N}$ approximation. 
I managed to get:
$$\frac{M^M}{(M-N)^{M-N}}\times \frac{1}{N^N}$$
but I suspect this is deviating from the main way of reaching the approximation. 
How is the approximation achieved?
 A: $$
\frac{M !}{\left(M-N\right)!}
~=~
\begin{alignat}{10}
M
& \times & \left(M - 1\right)
& \times & \left(M - 2\right)
& \times & ~\cdots~
& \times & \left(M - N + 1\right)
& \times & \left(M - N\right)
& \times & \left(M - N - 1\right)
% & \times & \left(M - N - 2\right)
& \times & ~\cdots~
% & \times & 2
& \times & 1
\\[-25px] \hline
& &
& &
& &
& &
& & \left(M - N\right)
& \times & \left(M - N - 1\right)
% & \times & \left(M - N - 2\right)
& \times & ~\cdots~
% & \times & 2
& \times & 1
\end{alignat}
$$
$$
\begin{align}
&~=~
M
\times \left(M - 1\right)
\times \left(M - 2\right)
\times ~\cdots~
\times \left(M - N + 1\right) \\[10px]
% &~=~ \prod_{i=M-N+1}^{M}{i}
\end{align}
$$
Since $\frac{N}{M} \ll 1 ,$ then $M \gg N,$ and $M \gg 1,$ so
$$
\begin{align}
\frac{M !}{\left(M-N\right)!}
&~=~
M
\times \underbrace{\left(M - 1\right)}_{\approx M}
\times \underbrace{\left(M - 2\right)}_{\approx M}
\times ~\cdots~
\times \underbrace{\left(M - N + 1\right)}_{\approx M} \\[10px]
&~\approx~
\underbrace{M \times \cdots \times M}_{N~\text{times}} \\[10px]
&~=~
M^N
\,,
\end{align}$$
so
$$
\frac{M !}{\left(M-N\right)!}
~\approx~
M^N
\,.$$
A: $$M! = M \times (M-1) \cdots \times (M-N+1) \times (M-N)!$$
$$\frac{M!}{(M-N)!}=\frac{M \times (M-1) \cdots \times (M-N+1) \times (M-N)!}{(M-N)!}$$
The terms $(M-N)!$ cancel,
$$\frac{M!}{(M-N)!}=\prod\limits_{i=0}^{N-1}\,(M-i)$$
We know $i < N$. Assuming $N \ll M$, we get $i \ll M$ and we can write $(M-i)$ as $M$, then
$$\frac{M!}{(M-N)!}=\prod\limits_{i=0}^{N-1}\,M = M^N.$$
Sorry, my initial response was wrong.
A: Just to see how to do it using Stirling's formula:
\begin{align}
\frac{M!}{(M- N)!} &= \frac{M^M e^{-M}}{(M-N)^{M-N} e^{-(M-N)}} \\
&= \frac{M^N}{e^N} \left( \frac{M}{M- N} \right)^{M-N} \\
&= \frac{M^N}{e^N} \left( 1 - \frac{N}{M-N} \right)^{-(M-N)} 
\end{align} 
But since $N \gg M$, we have $x \equiv (M-N)/N \gg 1$ and so
$$
\left( 1 - \frac{1}{x} \right)^{-Nx} = \left[ \left( 1 - \frac{1}{x} \right)^{x}\right]^{-N} \approx e^{N}
$$
using the approximation $(1 - 1/x)^x \approx e^{-1}$ for $x \gg 1$.  Thus,
$$
\frac{M!}{(M- N)!} \approx M^N.
$$
More generally, you can use Stirling's formula in the form $\ln n! \approx n \ln n - n$ to calculate corrections to this approximation.  If I've done my derivation correctly, for example, it turns out that 
$$
\ln \left[ \frac{M!}{(M- N)!} \right] \approx N \ln M + \mathcal{O} \left(\frac{N}{M}\right)^2.
$$
A: When density is low one could in principle follow the particles for a while. Assume the particles are distinguishable. Suppose there is one particle and that there are six sites: Ω = 6. Now two particles, a red one and a blue one - Ω = 36. Etcetera.
Interesting to note though that this approximation fails if there is a small hole in the partition. What final distribution would maximize Ω?
