Here's how I think about this. Suppose you're given $n$ terms $a_1,\dots, a_n$. Define $b_i = a_i / (i-1)\cdots(i-n)$, where the denominator skips the factor $(i-i)$. Consider the function
$$f(x) = b_1(x-2)(x-3)\cdots(x-n) + (x-1)b_2(x-3)\cdots(x-n) + \cdots + (x-1)\cdots(x-n+1)b_n + (x-1)\cdots(x-n)c$$
Then for $i$ an integer between $1$ and $n$, $f(i) = b_i \ast (i-1) \ast \cdots \ast (i-n)$, except that you skip the factor $(i-i)$. Thus, $f(i) = a_i$. But by changing $c$, you can make the next term $f(n+1)$ whatever you want.
