Algebraic proof for the following identity Give an algebraic proof that $\binom{n+1}{m+1} = \sum_{k=m}^{n} \binom{k}{m}$.
I've tried using Pascal's rule and looking for a telescopic sum, but I can't find one.
Any help is appreciated.

By the Binomial Theorem, we have
$$
\begin{align}
\color{#C00000}{\sum_{k=0}^n(1+x)^k}
&=\sum_{k=0}^n\sum_{j=0}^k\binom{k}{j}x^j\\
&=\sum_{j=0}^n\color{#00A000}{\sum_{k=j}^n\binom{k}{j}}x^j
\end{align}
$$
By the formula for the sum of a geometric series, we also have
$$
\begin{align}
\color{#C00000}{\frac{(1+x)^{n+1}-1}{x}}
&=\sum_{j=0}^n\color{#00A000}{\binom{n+1}{j+1}}x^j
\end{align}
$$
