Help understand second step in proof of ${n\choose m-1}+{n\choose m}={n+1\choose m}$ I have problem understanding second step in this proof:
\begin{align}
\binom n{m-1}+\binom nm&=\frac{n!}{(m-1)!(n-m+1)!}+\frac{n!}{m!(n-m)!}\\
&=\frac{n!m+n!(n-m+1)}{m!(n-m+1)!}\\
&=\frac{n!(n-m++1+m)}{m!(n-m+1)!}\\
&=\frac{n!(n+1)}{m!(n-m+1)!}\\
&=\frac{(n+1)!}{m!(n-m+1)!}=\binom{n+1}m
\end{align}
Why is it not like this:
\begin{align}
&=\frac{n!m!(n-m)!+n!(m-1)!(n-m+1)!}{(m-1)!(n-m+1)!m!(n-m)!}\\
\end{align}
But like this:
\begin{align}
&=\frac{n!m+n!(n-m+1)}{m!(n-m+1)!}\\
\end{align}
Does it just skip some step/steps that are clear to people with more mature math skills? I am doing exercises from Serge Lang's Basic mathematics as self study. Proof in question is from here: Binomial coefficient proof for ${n\choose m-1}+{n\choose m}={n+1\choose m}$
 A: The skipped step looks as following
$$
\begin{align}
\binom n{m-1}+\binom nm&=\frac{n!}{(m-1)!(n-m+1)!}+\frac{n!}{m!(n-m)!}\\
&\color{blue}{=\frac{n!m}{(m-1)!m(n-m+1)!}+\frac{n!(n-m+1)}{m!(n-m)!(n-m+1)}}\\
&=\frac{n!m}{m!(n-m+1)!}+\frac{n!(n-m+1)}{m!(n-m+1)!}.\\
\end{align}
$$
A: They do skip an intermediate step or two maybe. You just have to note that $m!=m(m-1)!$ and similarly $(n-m+1)! = (n-m+1)(n-m)!$. This allows you rewrite both terms on the right in the first line to expressions which have $m!(n-m+1)!$ in the denominator. 
A: Using 
 m!=m(m-1)!
(n-m+1)!=(n-m+1)!(n-m)

In step 3 on first term I multiple and divide by m! and in second term I multiply and divide by (n-m+1)!
In Step 4 I have opened factorial as I mentioned in using part and the rest is simple LCM now. 
\begin{align}
\binom n{m-1}+\binom nm&=\frac{n!}{(m-1)!(n-m+1)!}+\frac{n!}{m!(n-m)!}\\
&=\frac{m!n!}{m!(m-1)!(n-m+1)!}+\frac{(n-m+1)!n!}{(n-m+1)!m!(n-m)!}\\
&=\frac{m(m-1)!n!}{m!(m-1)!(n-m+1)!}+\frac{(n-m+1)(n-m)!n!}{(n-m+1)!m!(n-m)!}\\
&=\frac{mn!}{m!(n-m+1)!}+\frac{(n-m+1)n!}{(n-m+1)!m!}\\
&=\frac{n!m+n!(n-m+1)}{m!(n-m+1)!}\\
&=\frac{n!(n-m++1+m)}{m!(n-m+1)!}\\
&=\frac{n!(n+1)}{m!(n-m+1)!}\\
&=\frac{(n+1)!}{m!(n-m+1)!}=\binom{n+1}m
\end{align}
A: A proof using combinatory (more elegant).
Let consider a group $n+1$ people. Then, $\binom{n+1}{m}$ is the number of groups of $m$ people that we can form. Let Frank a member of these $n+1$ people. Notice that to make a group with $m$ people, we can consider group where Frank belong (and there are $\binom{n}{m-1}$) and Group where Frank do not belong (and there are $\binom{n}{m}$). Finally,
$$\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}.$$
