Transitivity of Big-O Estimates/Sequences

Theorem
Let $(a_n)$, $(b_n)$ and $(c_n)$ be sequences of real or complex numbers.

Let $a_n = O(b_n)$ and $b_n = O(c_n)$, where $O$ denotes big-O notation.

Then $a_n = O(c_n)$.

Proof
Because $a_n = O(b_n)$, there exists $K\geq0$ and $n_0 \in\N$ such that $|a_n| \leq K \cdot |b_n|$ for $n\geq n_0$.

Because $b_n = O(c_n)$, there exists $L\geq0$ and $n_1 \in\N$ such that $|b_n| \leq L \cdot |c_n|$ for $n\geq n_1$.

Then $|a_n| \leq KL \cdot |c_n|$ for $n\geq \max(n_0, n_1)$.

Thus $a_n = O(c_n)$.