Q:

Identity related to binomial distribution?

While writing a (non-math) paper I came across the following apparent identity:
$N \cdot \mathop \sum \limits_{i = 1}^N \frac{1}{i}\left( {\begin{array}{*{20}{c}}
{N - 1}\\
{i - 1}
\end{array}} \right){p^{i - 1}}{\left( {1 - p} \right)^{N - i}} = \frac{{1 - {{\left( {1 - p} \right)}^N}}}{p}$
where $N$ is a positive integer and $p$ is a nonzero probability. Based on intuition and some manual checks, this looks like it should be true for all such $N$ and $p$. I can't prove this, and being mostly ignorant about math, I don't know how to learn what I need to prove this. I'd really appreciate anything helpful, whether a quick pointer in the right direction or the whole proof (or a proof or example that the two aren't identical).

Note also that ${1 - {\left( {1 - p} \right)}^N} = {{\sum\limits_{i = 1}^N {\left( {\begin{array}{*{20}{c}}
N\\
i
\end{array}} \right){p^i}{{\left( {1 - p} \right)}^{N - i}}} }}$
and that ${p = {1 - {\left( {1 - p} \right)}^1}}$
For background, see the current draft with relevant highlightings here.

A:

Some manipulation gives the desired result. Mostly, one has to note that
$$\frac{N}{i}\binom{N-1}{i-1}=\binom{N}{i}.$$
For $\binom{N-1}{i-1}=\frac{(N-1)!}{(i-1)!((N-1)-(i-1))!}=\frac{(N-1)!}{(i-1)!(N-i)!}.$
Multiply the top by $N$, the bottom by $i$, and we get $\frac{N!}{i!(N-i)!}$, which is just $\binom{N}{i}$.
So our sum is
$$\sum_{i=1}^N\binom{N}{i}p^{i-1}(1-p)^{N-i}.$$
Multiply the inside by $p$, and divide by $p$ on the outside. We get
$$\frac{1}{p}\sum_{i=1}^Np^i(1-p)^{N-i}.$$
You have written down enough facts to take it the rest of the way.
Our expression is equal to 
$$\frac{1}{p}\left(\sum_{i=0}^N\binom{N}{i}p^i(1-p)^{N-i} -\binom{N}{0}p^0(1-p)^N     \right).$$
The $\sum_{i=0}^N$ stuff is just the binomial expansion of $(p+(1-p))^N$, so it is equal to $1$. Or alternately it is the sum of the binomial probabilities, so it is $1$. Finally, the term $\binom{N}{0}p^0(1-p)^N$ is an awkward way of writing $(1-p)^N$.

