Show That a Power Of 17 Exists Between $2^{16},2^{17}$

$$\begin{align}&\left[\frac{17}{16}\right]^{\large 4\cdot\color{#c00}4}\!\! = \left[1+\frac{1}{\color{#0a0}{16}}\right]^{\large\color{#0a0}{16}}\!\!<\, \color{#0a0}e\, < \color{c00}2^{\large\color{#c00} 4}\\[.2em] \overset{\large (\ {\phantom{|_|}} )^{\LARGE 1/\color{#c00}4}\!\!\!}\Longrightarrow\ \ \ &\left[\frac{17}{16}\right]^{\large 4}\! <\, \color{c00}2,\ \ {\rm so}\,\ \ \bbox[6px,border:1px solid #c00]{17^{\large 4} <\ 2\cdot 16^{\large 4} =\, 2^{\large17}}\end{align}\qquad$$

You suspect (correctly) that $2^{16} < 17^4 < 2^{17}$ since $17=2^4+1$

So try $$17^4=(16+1)^4 $$ $$= 16^4 + 4\times 16^3 +6\times 16^2+4\times 16^1 +1\times 16^0$$ $$< 16^4+4\times 16^3 +6\times 16^3+4\times 16^3 +1\times 16^3$$ $$=16^4+15 \times 16^3 $$ $$< 2 \times 16^4 = 2^{17}$$

and clearly $2^{16} = 16^4 < 17^4 $

Consider the fairly general case of showing, for some positive real numbers $a$, $b$, $c$ and $d$, with $a \neq 1$ and $d \gt c$, that

$$\exists \; k \in \mathbb Z \; \text{such that} \; a^{c} \lt b^{k} \lt a^{d} \tag{1}\label{eq1A}$$

Taking logs, say natural ones, of all of the parts gives

$$c\ln a \lt k \ln b \lt d \ln a \implies \frac{c\ln a}{\ln b} \lt k \lt \frac{d\ln a}{\ln b} \tag{2}\label{eq2A}$$

Thus, you just need to see if there's an integer value $k$ between the lower & upper limit values. In your case, you have $a = 2$, $b = 17$, $c = 16$ and $d = 17$. Plugging these values into \eqref{eq2A} gives

$$\begin{equation}\begin{aligned} & \frac{16\ln 2}{\ln 17} \lt k \lt \frac{17\ln 2}{\ln 17} \\ & 3.914\ldots \lt k \lt 4.159\ldots \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

This shows that just $k = 4$ works, as you've found and the other answers have shown as well. However, this method shown here will work in any more general, difficult cases to fairly easily determine if there's any $k$ and, if so, which value(s) of $k$ will work.

Similar Isaac YIU Math Studio's answer, we could instead use hexadecimal.

$2^{16}=16^4=10000_{16}$

$2^{17}=2\cdot2^{16}=20000_{16}$

As done in Henry's answer, we can binomially expand (or really just multiply out $11^4$) to see we then have:

$17^4=11_{16}^4=14641_{16}$

which is clearly between $10000_{16}$ and $20000_{16}$.