Smaller-Bigger and Bigger-Smaller

$17^{16} = (16+1)^{16}$

Using binomial expansion, we get:

$\displaystyle \Rightarrow \displaystyle \sum_{r=0}^{16} {16 \choose r} 16^{16 - r} (1)^r$

$\displaystyle \Rightarrow {16 \choose 0} 16^{16} + {16 \choose 1} 16^{15} + {16\choose 2} 16^{14} + \ldots$

$\displaystyle \Rightarrow (1)16^{16} + (16)16^{15} + \frac{(16)(15)(16^{14})}{2} + \ldots$

$\displaystyle \Rightarrow (2)\times 16^{16} + \frac{(15)(16^{15})}{2} + \ldots$

As we can see, every term in the expansion after the first two terms is smaller than $\displaystyle 16^{16}$ . So, we conclude:

$\displaystyle (2)\times 16^{16} + \frac{(15)(16^{15})}{2} + \ldots < (16)\times 16^{16}$

$\displaystyle \Rightarrow (16+1)^{16} < 16^{17}$

$\displaystyle \Rightarrow \boxed{17^{16} < 16^{17} }$