Standard Binomial Coefficients
Use the Binomial Theorem
$$
\begin{align}
(1+x)^m(1+x)^n
&=\sum_{k=0}^m\binom{m}{k}x^k\sum_{j=0}^n\binom{n}{j}x^j\\
&=\sum_{k=0}^m\sum_{j=0}^n\binom{m}{k}\binom{n}{j}x^{k+j}\\
&=\sum_{k=0}^m\sum_{r=k}^{k+n}\binom{m}{k}\binom{n}{r-k}x^r&&j=r-k\\
&=\sum_{r=0}^{m+n}\color{#C00000}{\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}}x^r\tag{1}\\
(1+x)^{m+n}
&=\sum_{r=0}^{m+n}\color{#C00000}{\binom{m+n}{r}}x^r\tag{2}
\end{align}
$$
Compare coefficients of $x^r$ in $(1)$ and $(2)$.
Generalized Binomial Coefficients
As Michael Hardy mentions, the formula is true, even when $m$ and $n$ are not integers. The binomial coefficients can be generalized to any real number in the top argument:
$$
\binom{x}{k}=\frac{x(x-1)(x-2)\dots(x-k+1)}{k!}\tag{3}
$$
When $x$ is a negative integer, $(3)$ gives the formula for the negative binomial coefficients:
$$
\begin{align}
\binom{-n}{k}
&=\frac{-n(-n-1)(-n-2)\dots(-n-k+1)}{k!}\\
&=(-1)^k\frac{(n+k-1)(n+k-2)(n+k-3)\dots n}{k!}\\
&=(-1)^k\binom{n+k-1}{k}\tag{4}
\end{align}
$$
The Generalized Binomial Theorem states that for any real $m$,
$$
(1+x)^m=\sum_{k=0}^\infty\binom{m}{k}x^k\tag{5}
$$
Note that when $m$ is a non-negative integer, $\binom{m}{k}=0$ for $k\gt m$ and so $(5)$ is a polynomial in that case.
Using $(5)$, we can imitate the proof for the standard binomial coefficients:
$$
\begin{align}
(1+x)^m(1+x)^n
&=\sum_{k=0}^\infty\binom{m}{k}x^k\sum_{j=0}^\infty\binom{n}{j}x^j\\
&=\sum_{k=0}^\infty\sum_{j=0}^\infty\binom{m}{k}\binom{n}{j}x^{k+j}\\
&=\sum_{k=0}^\infty\sum_{r=k}^\infty\binom{m}{k}\binom{n}{r-k}x^r&&j=r-k\\
&=\sum_{r=0}^\infty\color{#C00000}{\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}}x^r\tag{6}\\
(1+x)^{m+n}
&=\sum_{r=0}^\infty\color{#C00000}{\binom{m+n}{r}}x^r\tag{7}
\end{align}
$$
Compare coefficients of $x^r$ in $(6)$ and $(7)$.