Question: Let $p$ be an odd prime. Is it true that
$$-\left(\begin{array}{c}p+1\\ 2\end{array}\right)+\left(\begin{array}{c}p+1\\ 3\end{array}\right)\dots+\left(\begin{array}{c}p+1\\ p\end{array}\right)-\left(\begin{array}{c}p+1\\ p+1\end{array}\right)=p?$$

I could show that it is true for $p\in\{3,5\}$, but I could not prove it. 
Any help will be greatly appreciated!

Note: For positive integers $k$ and $n$, where $k\leq n$, 
we define $\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!(n-k)!}$.

Answer: 
It isn’t true: for $p=3$ you have $-\binom42+\binom43=-6+4=-2$. You need to switch the signs, so that the first term is positive. Then it’s true much more generally.

For any $n>0$ we have

$$\sum\_{k=0}^n(-1)^k\binom{n}k=0\;;$$

you see this by recognizing the summation as the expanion of $\big((-1)+1\big)^n$ using the binomial theorem (and in various other ways as well). After you reverse your signs, what you have (when $n=p+1$) is

$$\begin{align\*}
\sum\_{k=2}^n(-1)^k\binom{n}k&=\sum\_{k=0}^n(-1)^k\binom{n}k-\sum\_{k=0}^1(-1)^k\binom{n}k\\
&=0-\binom{n}0+\binom{n}1\\
&=n-1\;.
\end{align\*}$$
$$\sum\_{i=0}^{p+1}\binom{p+1}{i}(-1)^{i+1}=0$$ 

as $\binom{p+1}{i}=\binom{p+1}{i+2(i-\frac{p+1}{2})}\hspace{10mm}\forall i<\frac{p+1}{2}$ as the binom is symmetric and $p+1$ is even.

From there you just take

$$\sum\_{i=0}^{p+1}\binom{p+1}{i}(-1)^{i+1} + \binom{p+1}{1}-\binom{p+1}{0}=0+p+1-1=p$$

to eliminate the first 2 elements

You just need to reverse the signs to do this