<s>[INST] Solving a recurrence (with the form of a convolution) involving binomial coefficients While dealing with a problem related to intersection of hyperplanes I have come across the following recurrence to obtain the values of $K_{j}$
\begin{array}{cccccccccc}
1 & = & K_{1}\tbinom{n+1}{0}\\
1 & = & K_{1}\tbinom{n+2}{1} & +K_{2}\tbinom{n+2}{0}\\
1 & = & K_{1}\tbinom{n+3}{2} & +K_{2}\tbinom{n+3}{1} & +K_{3}\tbinom{n+3}{0}\\
1 & = & K_{1}\tbinom{n+4}{3} & +K_{2}\tbinom{n+4}{2} & +K_{3}\tbinom{n+4}{1} & +K_{4}\tbinom{n+4}{0}\\
 & \vdots\\
1 & = & K_{1}\tbinom{n+j}{j-1} & +K_{2}\tbinom{n+j}{j-2} & +K_{3}\tbinom{n+j}{j-3} & +K_{4}\tbinom{n+j}{j-4} & +\cdots & +K_{j-2}\tbinom{n+j}{2} & +K_{j-1}\tbinom{n+j}{1} & +K_{j}\tbinom{n+j}{0}
\end{array}
That is, the convolution 
$$1=\sum_{i=1}^{j}K_{i}\binom{n+j}{j-i}$$
or said otherwise, if we consider the sequence of coefficients, $K_{1},K_{2},K_{3},\ldots$
  the j
 th term is given by the recurrence

$$K_{j}=1-\sum_{i=1}^{j-1}K_{i}\binom{n+j}{j-i}$$

A direct calculation gives the following solutions
\begin{array}{cccc}
K_{1}= & +1 & = & +1\\
K_{2}= & -{n+1 \choose 1} & = & -\left({n+1 \choose 1}\right)\\
K_{3}= & +{n+2 \choose 2} & = & +\left({n+1 \choose 2}\right)\\
K_{4}= & -{n+3 \choose 3} & = & -\left({n+1 \choose 3}\right)\\
K_{5}= & +{n+4 \choose 4} & = & +\left({n+1 \choose 4}\right)\\
\vdots & \vdots &  & \vdots\\
K_{j}= & (-1)^{j-1}{n+j-1 \choose j-1} & = & (-1)^{j-1}\left({n+1 \choose j-1}\right)
\end{array}
where ${n \choose i}=\frac{n(n-1)\cdots(n-i+1)}{i!}$
  and $\left({N \choose i}\right)=\frac{n(n+1)\cdots(n+i-1)}{i!}$
However, this proves nothing.

Is it possible to derive a solution from the initial recurrence (convolution)?

 [/INST] We start from the convolution $$1=\sum_{i=1}^j K_i \binom{n+j}{j-i}$$ then multiply both sides by $x^{j-1}$ and sum from $j=1$ to $\infty$. For the RHS we obtain 
$$\sum_{j=1}^\infty \sum_{i=1}^j K_i \binom{n+j}{j-i}x^{j-1}=\sum_{i=1}^\infty \sum_{j=i}^\infty K_i \binom{n+j}{j-i}x^{j-1}$$ where we have reversed the order of summation.
The inner sum over $j$ may then be rewritten as 
$$\sum_{j=i} \binom{n+j}{j-i}x^{j-1}=\sum_{j=0}^\infty\binom{n+i+j}{j}x^{j+i-1} 
=\frac{x^{i-1}}{(1-x)^{n+i+1}}$$ where we have recognized the negative exponent binomial series $(1-x)^{-n-1}=\sum_{k=0}^\infty\binom{n+k}{k} x^k$. Completing the outer summation over $i$ then gives the RHS as
$$ \sum_{i=1}^\infty K_i \frac{x^{i-1}}{(1-x)^{n+i+1}}=\frac{1}{x (1-x)^{n+1}}\sum_{i=1}^\infty K_i \left(\frac{x}{1-x}\right)^i.$$ But the LHS is simply the geometric series $$\sum_{j=1}^\infty x^{j-1}=1+x+x^2+\cdots=\frac{1}{1-x},$$ and so we may write $$\sum_{i=1}^\infty K_i \left(\frac{x}{1-x}\right)^i=x(1-x)^n.$$ The substitution $x\mapsto x(1+x)^{-1}$ then brings this to the form
$$\sum_{i=0}^\infty K_{i+1} x^{i+1}=\frac{x}{(1+x)^{n+1}}=\sum_{i=0}^\infty \binom{n+i}{i} (-1)^i x^{i+1}$$ using the same binomial series as earlier. Identifying coefficients then finally yields  $$K_i = (-1)^{i-1}\binom{n+i-1}{i-1}=(-1)^{i-1}\frac{(n+1)(n+2)\cdots(n+i-1)}{(i-1)!}$$ which is equivalent to the formula deduced in the question statement.
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