Binomial theorem:
[tex](x+1)^{16}=\displaystyle\sum_{k=0}^{16}\dbinom{16}kx^{16-k}1^k=\sum_{k=0}^{16}\dbinom{16}kx^{16-k}[/tex]
The [tex]x^5[/tex] term corresponds with the term in the sum for which [tex]16-k=5\implies k=11[/tex]. For this [tex]k[/tex], we get
[tex]\dbinom{16}{11}x^{16-11}=\dbinom{16}{11}x^5[/tex]
So the coefficient you want to find is
[tex]\dbinom{16}{11}=\dfrac{16!}{11!(16-11)!}=4368[/tex]
