How is this a property of Pascal's triangle? For all non-negative integers $k$ and $n$,
$$
\dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n}{k} = \dbinom{n +1}{k+1}
$$
How is this a property of Pascal's triangle? I do not see at all how it relates. I'm looking for an explanation, as opposed to some calculation with algebraic manipulations.

By induction on $n$. If $n<k$ then the left hand side is an empty sum, and the right hand side has $0<n+1<k+1$ so $\binom{n+1}{k+1}=0$. For $n\geq k$ assume the result for $n-1$, so one has
$$
  \dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n-1}{k} =  \dbinom n{k+1}.
$$
Substituting that into the equation to be proved, we need to show $\binom n{k+1}+\binom n{k\rlap{\phantom+}}=\binom{n+1}{k+1}$, which is a familiar identity.
There is also a direct combinatorial interpretation of the formula. If you choose a subset of $k+1$ from the $n+1$ numbers $\{0,1,\ldots,n\}$, the last selected number $i$ satisfies $k\leq i\leq n$, and the remaining $k$ numbers can be chosen in $\tbinom ik$ ways among $\{0,\ldots,i-1\}$. Every subset is accounted for exactly once, so
$$\sum_{i=k}^n\binom ik=\binom{n+1}{k+1}.
$$
