The identity $$\binom{x+1}{k}-\binom{x}{k}=\binom{x}{k-1}$$ is claimed to hold (using the binomial polynomials, considered as lying in $\mathbf{Q}[x]$) for $k$ at least $1$. Proof: by the usual Pascal identity it holds for all $x=n$ at least $k$ and hence (using that these are polys of the same degree?) we can conclude (?) that these must be equal as polynomials.
Is my proof ok?
On
To show:
$$\binom{x+1}{k}-\binom{x}{k}=\binom{x}{k-1}$$
In factorial form, we have
$$\frac {(x+1)!}{k!(x+1-k)!}-\frac {x!}{k!(x-k)!}=\frac {x!}{(k-1)!(x-k+1)!}$$
Working with just the LHS,
$$\begin{align}\frac {(x+1)!}{k!(x+1-k)!}-\frac{x-k+1}{x-k+1}\frac {x!}{k!(x-k)!}&=\frac {x!}{(k-1)!(x-k+1)!}\\ \frac {(x+1)x!}{k!(x+1-k)!}-\frac {(x-k+1)x!}{k!(x-k+1)!}&=\frac {x!}{(k-1)!(x-k+1)!}\\ \frac {kx!}{k!(x+1-k)!}&=\frac {x!}{(k-1)!(x-k+1)!}\\ \frac {x!}{(k-1)!(x+1-k)!}&=\frac {x!}{(k-1)!(x-k+1)!}\end{align}$$
Not messy at all... </tongue-in-cheek>.
Your idea goes through nicely. Let $$P(x)=\binom{x+1}{k}-\binom{x}{k}-\binom{x}{k-1}.$$ Then $P(x)$ is the zero polynomial or a polynomial of degree $\le k$.
By the usual combinatorial argument, we have $P(x)=0$ for $x=k, k+1, \dots, 2k$. Since a non-zero polynomial of degree $\le k$ over a field has at most $k$ zeros, it follows that $P(x)$ is the zero polynomial.
More briefly, the equation $P(x)=0$ has infinitely many zeros, so $P(x)$ is the zero polynomial.