Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$ 
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
  $$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}
$$

My Attempt:
For $(\mathrm{i})$, We can write $\binom{n}{m}$ as the coefficient of $x^m$ in $(1+x)^n$. Thus we can also write $\binom{n-1}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-1}$ and $\binom{n-2}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-2}$. So we have to find the coefficient of $x^m$ in
$$
(1+x)^n+(1+x)^{n-1}+(1+x)^{n-2}+........+(1+x)^{m}
$$
Using the formula for the sum of a geometric progression, this sum equals
$$\frac{(1+x)^{n+1}-(1+x)^m}{x}$$
So we now need to find the coefficient of $x^m$ in
$$
\frac{(1+x)^{n+1}-(1+x)^m}{x}
$$
or, equivalently, we need to find the coefficient of $x^{m+1}$ in
$$
(1+x)^{n+1}-(1+x)^m = \binom{n+1}{m+1}
$$
We can use a similar method to solve $(\mathrm{ii})$. Can these questions be solved using combinatorial methods instead?

First problem: We have $n+1$ different doughnuts (labelled $1$ to $n+1$) lined up in a row, and want to choose $m+1$ of them for breakfast. This can be done in $\binom{n+1}{m+1}$ ways. Let us count another way. 
Maybe the leftmost doughnut chosen is $1$. There are $\binom{n}{m}$ ways to choose the rest.
Maybe the leftmost doughnut chosen is $2$. There are then $\binom{n-1}{m}$ ways to choose the others.
And so on. 
Second Problem: This is done similarly. This time we are choosing $m+2$ doughnuts from $n+2$ doughnuts, labelled $1$ to $n+2$ and lined up in that order.
Look at the leftmost two doughnuts chosen. If the second one is Doughnut $2$, there are $\binom{n}{m}$ ways to choose the rest. 
If the second one chosen is Doughnut $3$, there are $2$ ways to choose the first one, and $\binom{n-1}{m}$ ways to choose the rest, for a total of $2\binom{n-1}{2}$.
If the second one chosen is Doughnut $4$, there are $3$ ways to choose the first one, and $\binom{n-2}{m}$ ways to choose the rest, for a total of $3\binom{n-2}{m}$.
And so on.
