One easy NP-complete problem collapses P and NP

Suppose $B$ is NP-complete and there is an algorithm solving $B$ in time $O(n^{k_1})$. Take any $A \in NP$; by NP-completeness of $B$, $A \le_p B$, meaning there is a reduction function $f$ computable in time $O(n^{k_2})$ that maps instances of $A$ to instances of $B$ while preserving yes/no answers.

On an input $x$ of length $n$ for $A$, first compute $f(x)$: this takes time $O(n^{k_2})$, and in particular the output $f(x)$ has length at most $O(n^{k_2})$ (a polynomial-time algorithm cannot write more output than time it runs). Then run the polynomial-time algorithm for $B$ on $f(x)$: since $|f(x)| = O(n^{k_2})$, this takes time $O\big((n^{k_2})^{k_1}\big) = O(n^{k_1 k_2})$.

The total running time is $O(n^{k_2}) + O(n^{k_1 k_2}) = O(n^{k_1 k_2})$, still a polynomial in $n$ (the composition of two polynomials is a polynomial). By the correctness of the reduction, $x$ is a yes-instance of $A$ if and only if $f(x)$ is a yes-instance of $B$, so this combined procedure correctly decides $A$ in polynomial time.

Since $A \in NP$ was arbitrary, every NP problem has a polynomial-time algorithm, i.e. $NP \subseteq P$. Combined with the always-true containment $P \subseteq NP$, this gives $P = NP$.