Perhaps the simplest way to compute the sum would just be to evaluate each term in the sum and take the total:
[tex]\displaystyle \sum_{k=0}^7 \frac{15!}{(15-k)!k!} = \frac{15!}{15!0!} + \frac{15!}{14!1!} + \frac{15!}{13!2!} + \cdots + \frac{15!}{8!7!}[/tex]
Then you have
15! / (15! 0!) = 1
15! / (14! 1!) = 15
15! / (13! 2!) = 105
and so on. Adding these results together gives 16,384.
I think a better solution would be to make use of the symmetry of the binomial coefficients, captured by the identity
[tex]\displaystyle \binom nk = \binom n{n-k}[/tex]
where [tex]\binom nk = \frac{n!}{(n-k)!k!}[/tex].
By the binomial theorem, we have
[tex]\displaystyle \sum_{k=0}^{15} \frac{15!}{(15-k)!k!} = \sum_{k=0}^{15} \binom{15}k = \sum_{k=0}^{15} \binom{15}k1^k1^{15-k} = (1+1)^{15} = 2^{15}[/tex]
Split up the "complete" sum at k = 7:
[tex]\displaystyle \sum_{k=0}^7 \binom{15}k + \sum_{k=8}^{15}\binom{15}k = 2^{15}[/tex]
Use the identity above to rewrite the second sum:
[tex]\displaystyle \sum_{k=0}^7 \binom{15}k + \sum_{k=8}^{15}\binom{15}{15-k} = 2^{15}[/tex]
Shift the index on the second sum to make it start at k = 0, and the result follows:
[tex]\displaystyle \sum_{k=0}^7 \binom{15}k + \sum_{k=7}^{0}\binom{15}{15 - (15 - k)} = 2^{15} \\\\ \sum_{k=0}^7 \binom{15}k + \sum_{k=7}^{0}\binom{15}k = 2^{15} \\\\ \sum_{k=0}^7 \binom{15}k + \sum_{k=0}^7\binom{15}k = 2^{15} \\\\ 2\sum_{k=0}^7 \binom{15}k = 2^{15} \\\\ \sum_{k=0}^7 \binom{15}k = 2^{14} = \boxed{16,384}[/tex]
