$\binom{n}{n+1} = 0$, right? I was looking at the identity $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}, 1 \leq r \leq n$, so in the case $r = n$ we have $\binom{n}{n} = \binom{n-1}{n-1} + \binom{n-1}{n}$ that is $1 = 1 + \binom{n-1}{n}$ thus $\binom{n-1}{n} = 0$, am I right? 

In my small treatize about "sums of like powers" (pg 14 ff) where I "re-invented" the integrals of the Bernoulli polynomials by a matrix-approach, I arrived at a formulation, where setting $ x=\binom{n}{n+1}$  to $ {1 \over (n+1)*(-1)!} \ne 0 $ was a meaningful thing, because it appeared in a formalism, where writing $ \frac 11b + \frac 12b + \frac13b + ... = (\frac 11 + \frac 12 + \frac 13 + ...)b = \frac {\zeta(1)}b = -1$ (where $b = \frac 1{(-1)!} $) allowed to arrive at the correct solution for the Bernoulli-/Zeta-polynomials. (We can argue, that $\lim _{\delta \to 0} { \zeta(1+\delta) \over \Gamma(\delta) } = 1$ and $\lim _{\delta \to 0} { \zeta(1-\delta) \over \Gamma(-\delta) } = -1$)    
This is not a proof, that $ x=\binom{n}{n+1}$ need not always "be taken as zero", but it made me seriously thinking, how (and where else) such a setting actually makes sense.
