I know that this problem might qualify as a duplicate, but I didn't find any satisfactory answers among the similar posts.
We want to show by induction that \begin{equation} \sum_{i=0}^{n} \binom{i}{m} = \binom{n+1}{m+1} \end{equation} for all $n \geq 0$ and $m \geq 0$. In the basis step, we put $n=1$ and get \begin{align*} &\sum_{i=0}^1 \binom{i}{m} = \binom{0}{m} + \binom{1}{m} \end{align*} We see that $\binom{0}{m} = 0$ if $m = 0$ and $\binom{0}{m} = 1$ if $m > 0$. Lets first look at the case $m=0$: \begin{align*} &\sum_{i=0}^{1} \binom{i}{0} = 1 + \frac{1!}{0!1!} = 2 = \binom{2}{1} \end{align*} Thus, the equation holds for $n=1$ when $m = 0$. When $m>0$ we get \begin{align*} &\sum_{i=0}^{1} \binom{i}{m} = \binom{1}{m} = \frac{1!}{m!(1-m)!} \end{align*}
I don't know where to go from here. I want to end up with $\frac{2!}{(1+m)!(1 - m)!}$, so I tried to multiply with $\frac{(m+1)!}{(m+1)!}$, but that just got me here: \begin{align*} \sum_{i=0}^{1} \binom{i}{m}&= \frac{1}{m!(1-m)!} \cdot \frac{(m+1)!}{(m+1)!} \\ &= \frac{(m+1)}{(1-m)!(m+1)!} \end{align*}
I didn't get anywhere using Pascals rule, either. And this is just the basis step of the induction. The induction step, with $n = k+1$, seems just as difficult to me.