Binomial identity in a Change of Basis Matrix I need the following identity in my research and have a rather ugly proof but cannot find anything more elegant. 
If $L\colon\mathbb{Q}[m]\to\mathbb{Q}[m]$ is a linear map which satisfies
$L\left(\binom{m}{c+d}\right)=L\left(\binom{m}{c}\right)L\left(\binom{m}{d}\right)$
for all integer $c,d\geq0$, then $L$ also satisfies
$L\left(\binom{m+a+b}{c+d}\right)=L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)$
for all integer $a,b,c,d\geq0$.
We shall proceed by induction on $a+b$.
For the base case where $a+b=0$, $a=b=0$ and the result is a re-indexed form of the assumption on $L$.
Now suppose that the result holds for all $a+b=n$ and that $a+b=n+1$.
Without loss of generality, $a\geq1$.
Then by Pascal's identity and the inductive assumption,
\begin{align*}
L\left(\binom{m+a+b}{c+d}\right)&=L\left(\binom{m+(a-1)+b}{(c-1)+d}\right)+L\left(\binom{m+(a-1)+b}{c+d}\right)\\
&=L\left(\binom{m+(a-1)}{c-1}\right)L\left(\binom{m+b}{d}\right)+L\left(\binom{m+(a-1)}{c}\right)L\left(\binom{m+b}{d}\right)\\
&=L\left(\binom{m+(a-1)}{c-1}+\binom{m+(a-1)}{c}\right)L\left(\binom{m+b}{d}\right)\\
&=L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)
\end{align*}
which shows the inductive step and completes the proof. Does anyone have a better proof of this? Furthermore, is this the sort of result one would include in a paper or exclude as obvious enough for the reader to conclude?

Vandermonde's identity and diagonal summation can be used to give a non-inductive proof:
\begin{align*}
L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)&=L\left(\sum_{i=0}^c\binom{m}{i}\binom{a}{c-i}\right)L\left(\sum_{j=0}^d\binom{m}{j}\binom{b}{d-j}\right)\\
&=\sum_{i=0}^c\sum_{j=0}^d\binom{a}{c-i}\binom{b}{d-j}L\left(\binom{m}{i}\right)L\left(\binom{m}{j}\right)\\
&=\sum_{k=0}^{c+d}L\left(\binom{m}{k}\right)\sum_{i=0}^c\binom{a}{c-i}\binom{b}{d-(k-i)}\\
&=L\left(\sum_{k=0}^{c+d}\binom{m}{k}\binom{a+b}{c+d-k}\right)\\
&=L\left(\binom{m+a+b}{c+d}\right).
\end{align*}
