As part of a problem, I'm trying to find an estimation for the smallest $n$ (as a function of $k$) such that: $$\binom{n}{k}\biggl(1-\frac{1}{2^k}\biggr)^{(n-k)}<1$$
It's probably not possible to explicitly extract $n$ from this inequality, but I would be happy to get a good estimaion.
Theorem. Assume $k \ge 20$. Then your inequality is satisfied when $n = k + k^22^k$. However, if $k \le n \le k + k^22^{k-1}$, then the inequality does not hold. Therefore, for the smallest $n \ge k$ for which the inequality is satisfied, we have $k^22^{k-1} < n -k \le k^22^k$.
Proof. First we prove that for $n = k + k^22^k$, the inequality holds. Note that, for $k \ge 20$, we have $n = k + k^2 2^{k} < e^{k}$ and furthermore note that your inequality is equivalent to the following inequality:
\begin{equation} \binom{n}{k} < \left(\frac{2^k}{2^k-1}\right)^{(n-k)} \end{equation}
Now, since $n < e^k$, we get $\log(n) < k$ and therefore $k \log(n) < k^2 = \frac{(n-k)}{2^k}$.
Using $k \log(n) < \frac{(n-k)}{2^k}$ and the inequality $\frac{x}{1+x} < \log(1+x)$ (valid for all $x > 0$), it's a straight-forward calculation:
\begin{align*} \binom{n}{k} &< n^k \\ &= e^{k \log(n)} \\ &< e^{(n-k)\cdot\frac{1}{2^k}} \\ &< e^{(n-k)\cdot \log\left(\frac{2^k}{2^{k}-1}\right)} \\ &= \left(\frac{2^k}{2^k-1}\right)^{n-k} \end{align*}
For the other direction, first of all note that when $n = k$, it's clear. For $k+1 \le n \le k + 2^k$ we have, on the one hand, $\binom{n}{k} \ge k+1 \ge 4$ for $k \ge 3$, while on the other hand $\left(1 - \frac{1}{2^k} \right)^{(n-k)} \ge \left(1 - \frac{1}{2^k} \right)^{2^k} \ge 0.25$. We may therefore assume the existence of a constant $c$ with $2 < c \le k^2$ such that $n = k + c2^{k-1}$.
We now use the well-known inequalities $\binom{n}{k} > \frac{n^k}{k^k}$ and $x > \log(1 + x)$ to prove that the inequality $\binom{n}{k} > \left(\frac{2^k}{2^k-1}\right)^{(n-k)}$ holds. In order to show this, we also need the inequality $k(k-1)\log(2) - (k-2) \log(k) > k^22^{k-1}\frac{1}{2^k-1}$, which is valid for $k \ge 17$. We then have:
\begin{align*} \binom{n}{k} &> \frac{n^k}{k^k} \\ &> \frac{(n-k)^k}{k^k} \\ &= \frac{c^k 2^{k(k-1)}}{k^k} \\ &> e^{k(k-1)\log(2) - (k-2) \log(k)} \\ &> e^{c2^{k-1}\cdot\frac{1}{2^k-1}} \\ &> e^{(n-k)\cdot \log\left(\frac{2^k}{2^{k}-1}\right)} \\ &=\left(\frac{2^k}{2^k-1}\right)^{n-k} \end{align*}
And this finishes the proof. By being slightly more careful with estimates, I reckon you should be able to prove that the smallest $n$ is asymptotically equal to $\log(2)k^22^k$.