Question: Prove that the sequence $(x_n)_{n=1}^\infty$ defined by $x_n=\sqrt[3]{n}+1$ tends to infinity.
Definition: $\forall K \in \mathbb{R} \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:x_n>K $
Proof: Given any $K$ in real numbers, choose $N=(K-1)^3$. Given any $n$ in natural numbers with $n>N=\lceil(K-1)^3\rceil$ we have $x_n=\sqrt[3]{n}+1\geq \sqrt[3]{N}+1\ge K$
Is my proof correct?