Quick question about bound on certain function while computing limit

Question

I'm given the limit: $$\lim_{n\to\infty}\sqrt[n]{n^2+n^3+n^4+2^n+3^n+4^n}$$ The solution says: Because exponentials grow faster than polynomials, there is some $n_0$ such that $\forall n\in\Bbb{N},n\geq n_0:n^2\leq n^3\leq n^4 \leq 2^n \leq 3^n \leq 4^n$ so that $$\forall n\in\Bbb{N},n\geq n_0:4\leq\sqrt[n]{n^2+n^3+n^4+2^n+3^n+4^n}\leq\sqrt[n]{6\cdot4^n}=4\sqrt[n]{6}$$ and by squeeze lemma it follows that the limit is $4$. I'm quite confused, where did we get the bound that the function is always greater than $4$ i.e. the first inequality.

Answer 1

As is written in the comments we can see this through some straightforward replacements:

$$4$$ $$=\sqrt[n]{0+0+0+0+0+4^n}$$

$$\leq\sqrt[n]{n^2+n^3+n^4+2^n+3^n+4^n}$$ $$\leq\sqrt[n]{4^n+4^n+4^n+4^n+4^n+4^n}$$ $$=\sqrt[n]{6\cdot4^n}$$ $$=4\sqrt[n]{6}$$

Where for the last inequality we need $n>3$, since $n^4\leq 4^n$, fails for some small values of $n$, for $n=3$.