Let $n,m \in \mathbb{Z}^+$ and $(B(t))$ be a standard Brownian motion.
I know that by defintion $[B^n,B^m]_t$ is the mutual variation of $(B^n(t))$ and $(B^m(t))$ if for any sequence of partitions $(\Delta_n)$ with $|\Delta_n| \to 0$, we have \begin{align*} \sum_{k=1}^n \big( B^n(t_k) - B^n(t_{k-1}) \big) \big( B^m(t_k) - B^m(t_{k-1}) \big) \to [B^n,B^m]_t. \end{align*} in probability.
I think this question should be asked frequently (there might even be a diplicate somewhere), yet I couldn't find an answer anywhere or calculate it myself.
By Ito's formula, $$ B_t^n = n\int_0^t B^{n-1}_s\,dB_s+{n\choose 2}\int_0^t B_s^{n-2}\,ds, $$ and likewise for $B^m_t$. The covariation of $B^n$ with $B^m$ is therefore $$ mn\int_0^t B_s^{m+n-2}\,ds. $$