With minimal computation (i.e., without computing any of the actual quantities), show that $$\exists \; k \in \mathbb Z \; \text{such that} \; 2^{16} \lt 17^{k} \lt 2^{17}$$
My 17th birthday is coming up soon, and one of my friends at school sent me this question- she said that she had made it up for my birthday. I tried solving it by expanding $(2^{16}-1)+1=(2^{15}+2^{14}+2^{13}+...+2^{2}+2^{1}+2^{0})+1$ and $(2^{17}-1)+1=(2^{16}+2^{15}+2^{14}+...+2^{2}+2^{1}+2^{0})+1$, and seeing if "polynomializing" the powers of $2$ in this way helped find the power of $17$ somehow (I know from computation that the power is $17^{4}$).
I also noted that $17=2^{4}+1\gt2^{4}$, so $17^{4}\gt (2^{4})^{4}=2^{16}$, but I don't think I can perform the same sort of trick for the upper bound of $2^{17}$ without exactly the type of computation my friend wants me to avoid.
So my question is: can I actually solve my friend's problem without any sort of heavy calculation, or is she just messing with me? (Note: I would not be shocked if it were to be the latter, because that is exactly the sort of thing I have known her to do in the past.)
from Hot Weekly Questions - Mathematics Stack Exchange
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