It's possible to solve this equation without using Lambert function or any numerical method, but only with ordinary algebra?
$n^{k}lg_2(n) \le k^n$ with $k,n>0, k \in \mathbb{R}$
For $k=\frac{4}{3}$, the calculator says that the solution is $n \ge 22.78$, but I need to prove it, and I'm stuck applying properties of exponential and logarithmic functions.
I can squeeze you a small solution under a more restricted value of $k$. Perhaps you can use this to satisfy yourself in the general problem. I found it irritating to manage the case when $0 < k < 1$ but I am sure you can use my method below to whip something up.
For a given $k \geq 1$, and assume $n \geq 1$, we have $k \leq k^n$ so if we can satisfy $n^k \log_2(n) \leq k$ we can definitely satisfy the original inequality. Additionally, $n^k \log_2(n) \leq n^{k+1}$ whenever $\log_2(n) \leq n$ (which is always when $n > 0$). So if we can solve $n^{k+1} \leq k$ then we can definitely solve the original inequality. If we take the base-$k$ logarithm of both sides we get $(k+1) \log_k(n) \leq 1$, or
$$\log_k(n) \leq \frac{1}{k+1} \\ n \leq k^{\frac{1}{k+1}}$$
For instance, if $k=\frac{4}{3}$ then $\frac{4}{3}^{\frac{3}{4}} > 1.2$ and we can set $n = \frac{6}{5}$.