Question: The question is let $\{m\_1, m\_2, m\_3, \dots \}$ be a sequence of numbers where $m\_k\geq 0$ for every $k \geq 1$.

Let $$M\_n = \sum\_{k=1}^n m\_k $$ when $n \geq 1$ is an integer.

Show that if $$\prod\_{k=1}^{n-1}{(M\_k+1)\choose(M\_k)}= \frac{(m\_1+m\_2+\cdots+m\_n)!}{(m\_1!m\_2!\cdots m\_n!)}$$

When I try I get an expansion that is for example the first term and I choose a random letter in this case 2 for $m\_1$ the number $2!+1\over 2!(2+1-2)!$.

Is this the correct way to expand this series?

Thanks in advance. Let me know if anything should be clarified.

Answer: 
[![enter image description here](https://i.stack.imgur.com/1bKrC.jpg)](https://i.stack.imgur.com/1bKrC.jpg)

You get $\cos(7x)=0$ and $\cos(3x)\sin(4x) \ne 0$.

As you've already figured out

$\cos(7x) = 0 \implies x = \dfrac{2j+1}{14}\pi \quad (j \in \mathbb Z)$

$\cos(3x) = 0 \implies x = \dfrac{2k+1}{6}\pi \quad (k \in \mathbb Z)$

$\sin(4x) = 0 \implies x = \dfrac{\ell}{4} \pi \quad (\ell \in \mathbb Z)$

We look for coincidence of points
\begin{align}
 \dfrac{2j+1}{14}\pi &= \dfrac{2k+1}{6}\pi \\
 6j+3 &= 14k+7 \\
 3j - 7k &= 2 \\
 \hline
 j &= 7n + 3\\
 k &= 3n + 1\\
\hline
 \theta &\ne \dfrac{2n+1}{2}\pi
\end{align}

\begin{align}
 \dfrac{2j+1}{14}\pi &= \dfrac{\ell}{4} \pi \\
 4j+2 &= 7 \ell \\
 7 \ell - 4j &= 2 \\
 \hline
 j &= 7n + 3\\
 \ell &= 4n + 2\\
\hline
 \theta &\ne \dfrac{2n+1}{2}\pi
\end{align}

So we have to require $j \ne 7n + 3$, which we can write as 
$j \not \equiv 3 \pmod 7$. In summary

$x = \dfrac{2j+1}{14}\pi\;$ where $\; j \not \equiv 3 \pmod 7$
If you can use the complex exponential definition of trig functions,

$$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
$$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$$
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{(e^{i\theta} - e^{-i\theta})}{i(e^{i\theta} + e^{-i\theta})}$$
$$\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{i(e^{i\theta} + e^{-i\theta})}{(e^{i\theta} - e^{-i\theta})}$$

Then your equation becomes:

$$\frac{(e^{i3x} - e^{-i3x})}{i(e^{i3x} + e^{-i3x})} = \frac{i(e^{i4x} + e^{-i4x})}{(e^{i4x} - e^{-i4x})}$$
$$(e^{i3x} - e^{-i3x})(e^{i4x} - e^{-i4x}) = -(e^{i3x} + e^{-i3x})(e^{i4x} + e^{-i4x})$$
$$e^{i7x} - e^{-ix} - e^{ix} + e^{-i7x} = -(e^{i7x} + e^{-ix} + e^{ix} + e^{-i7x})$$
$$2e^{i7x} + 2e^{-i7x} = 0$$
$$2(2\cos(7x)) = 0$$
$$\cos(7x) = 0$$
$$7x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z}$$
$$x = \frac{\pi}{14} + \frac{\pi}{7}n$$