<s>[INST] When is $\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod {2^n} \ne 0\;$? Is
$$\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod{2^n} \ne 0\;$$
always true when $n \ge 3$.
Baker's theorem on transcendental numbers that provide bounds for diophantine equations may be useful, but I will leave that to the experts.
 [/INST] The following may be a starting point:
Write $7^n \mod 4^n$: 
$$ 7^n =a\cdot 4^n + b \quad b<4^n $$
$$ \left\lfloor \frac{7^n}{2^n} \right\rfloor \equiv \left\lfloor \frac b{2^n} \right\rfloor \pmod{2^n}$$
and this is $0$ iff $b<2^n$.
Well, $7^n=(2^3-1)^n=\displaystyle\sum_{0\le i\le n} \binom{n}{i} 2^{3i} \cdot (-1)^{n-i} $, and those with indices $3i\ge 2n$ vanish, thus
$$7^n \equiv (-1)^n\sum_{ i < 2n/3 } \binom ni (-8)^{i} = ... \pmod{4^n}$$
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