Summation Kills

Using repeatedly the Hockey Stick identity : $\Large\displaystyle\sum_{\color{#D61F06}{n_{60}}=0}^{2}\sum_{\color{#D61F06}{n_{59}}=0}^{n_{60}}\cdots\sum_{\color{#D61F06}{n_2}=0}^{n_3}\sum_{\color{#D61F06}{n_1}=0}^{n_2}\sum_{\color{#D61F06}{n_0}=0}^{n_1}(1)$ $\Large =\displaystyle\sum_{\color{#D61F06}{n_{60}}=0}^{2}\sum_{\color{#D61F06}{n_{59}}=0}^{n_{60}}\cdots\sum_{\color{#D61F06}{n_2}=0}^{n_3}\sum_{\color{#D61F06}{n_1}=0}^{n_2}\dbinom{n_1+1}1$ $\Large =\displaystyle\sum_{\color{#D61F06}{n_{60}}=0}^{2}\sum_{\color{#D61F06}{n_{59}}=0}^{n_{60}}\cdots\sum_{\color{#D61F06}{n_2}=0}^{n_3}\dbinom{n_2+2}{2}$ $\cdots\\\cdots$ $\Large =\displaystyle\sum_{\color{#D61F06}{n_{60}}=0}^{2}\sum_{\color{#D61F06}{n_{59}}=0}^{n_{60}}\dbinom{n_{59}+59}{59}$ $\Large =\displaystyle\sum_{\color{#D61F06}{n_{60}}=0}^{2}\binom{n_{60}+60}{60}=\binom{2+61}{61}$ $\Large\therefore \dbinom{63}{61}=\boxed{\color{#007fff}{1953}}$

If you're looking for a proof here are some cute proofs posted... :-)