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.

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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 $$7^n \equiv (-1)^n\sum_{ i < 2n/3 } \binom ni (-8)^{i} = ... \pmod{4^n}$$
One approach would be $7^n = (8 - 1)^n = (2^3 - 1)^n$. Doing binomial expansion, we get $7^n/2^n = 2^n x + k$ (k is sum of all binomial terms $i < n /3$). Little improved because it leaves less terms than above, but essentially same methodology.