Given is the double sum
$$\sum_{i = 0}^N {N\choose i}\alpha_i\sum_{j=0}^{N-i} {{N-i}\choose j}v^{(N-i-j)}u^{(j)}$$
where $$v^{(m)}$$ is the m-th derivative (single variable). I'd like to rearrange the first expression and arrive at
$$\sum_{l = 0}^N \beta_l v^{(l)}$$ The 'betas' can be anything that comes out from reworking the first expression. I am unsure how to do this.
With $l=N-i-j$, and thus $j=N-i-l$, you have
\begin{eqnarray*} && \sum_{i = 0}^N {N\choose i}\alpha_i\sum_{j=0}^{N-i} {{N-i}\choose j}v^{(N-i-j)}u^{(j)}\\&=& \sum_{i = 0}^N\sum_{j=0}^{N-i} {N\choose i}\alpha_i {{N-i}\choose j}v^{(N-i-j)}u^{(j)} \\&=& \sum_{i = 0}^N\sum_{l=0}^{N-i} {N\choose i}\alpha_i {{N-i}\choose N-i-l}v^{(l)}u^{(N-i-l)} \\&=& \sum_{i = 0}^N\sum_{l=0}^{N-i} {N\choose i}\alpha_i {{N-i}\choose l}v^{(l)}u^{(N-i-l)} \\&=& \sum_{l = 0}^N\sum_{i=0}^{N-l} {N\choose i}\alpha_i {{N-i}\choose l}v^{(l)}u^{(N-i-l)} \\&=& \sum_{l = 0}^Nv^{(l)}\sum_{i=0}^{N-l} {N\choose i}\alpha_i {{N-i}\choose l}u^{(N-i-l)}\;, \end{eqnarray*}
so
$$ \beta_l=\sum_{i=0}^{N-l} {N\choose i}\alpha_i {{N-i}\choose l}u^{(N-i-l)}\;. $$
There's a general pattern here that you can apply in transforming double sums of products: Pull the double sum to the front, relabel the indices (if required), switch the sums, and pull the factors independent of the second summation index in front of the second sum.