Is it valid to define $\binom{n}{n+k} = 0$

You may extend the definition of $n \choose r$, by making it 0 if r<0 or r>n.

If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.

It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $\binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.

The extended definition is \begin{align*} \binom{r}{k}= \begin{cases} \frac{r(r-1)\cdots (r-k+1)}{k!}&\qquad \text{integer }k\geq 0\\ 0&\qquad \text{integer }k <0 \end{cases}\tag{1} \end{align*}

This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.

From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $\binom{n}{n+k}=0$.