Proving $\binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n}$ algebraically I am working through the exercises and have spent half a day on one problem so I decided to get some help because I can't figure it out.
Show that if $n$ is a positive integer at most equal to $m$, then
$$
\binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n}
$$
I expand the sum of two binomial coefficients and get
$$
\frac{m!}{n!(m-n)!} + \frac{m!}{(n-1)!(m-n+1)!}
$$
Now, I have trouble summing up the two fractions, specifically, finding the common denominator of two fractions. I tried all sorts of things I could think of but can't seem to get to $n!(m-n+n)!$ which is in the solutions at the back of the book.
My last try
$$
\frac{m!(n!(m-n)!) + m!((n-1)!(m-n+1)!)}{n!(m-n)!(m-n+1)!(n-1)!} =
$$
$$
= \frac{m!(n!(m-n)! + (n-1)!(m-n+1)!)}{n!(m-n+1)!(m-n)!(n-1)!} =
$$
$$
= \frac{m!(mn!-nn!+((mn-n^2+n)!-(m-n+1)!)!}{n!(m-n+1)!(m-n)!(n-1)!}
$$
And so on, I feel like I am missing something as I don't see any opportunity to simplify this mess.

Proof using formula:
$$
{{m\choose n} := {m(m-1)\ldots(m-n+1)\over n!}}
$$
$$
\eqalign{{m\choose n}+{m\choose n-1}&=
{m(m-1)\ldots(m-n+1)\over n!}+{m(m-1)\ldots(m-(n-1)+1)\over (n-1)!}\cr 
&={m(m-1)\ldots(m-n+2)\cdot(m-n+1)\over n!}+{m(m-1)\ldots(m-n+2)\over (n-1)!} \cdot {n \over n}\cr 
&={m(m-1)\ldots(m-n+2)\cdot[(m-n+1)+n]\over n!}\cr 
&={m(m-1)\ldots(m-n+2)(m+1)\over n!}\cr 
&={(m+1)m(m-1)\ldots[(m+1)-n+1)]\over n!}\cr 
&={m+1\choose n}.
}
$$
