Coupled coils and coupling coefficient: from fraction of total flux to inductance definition

It is possible to relate the coupling coefficient $k$ to both the total (${\varphi }_1$ or ${\varphi }_2$) and shared flux (${\varphi }_{12}$ or ${\varphi }_{21}$) or to the self ($L_1$ and $L_2$) and mutual ($M$) inductances: $$k=\frac{M}{\sqrt{L_1{L}_2}}=\frac{{\varphi }_{12}}{{\varphi }_1}=\frac{{\varphi }_{21}}{{\varphi }_2}$$

Simple proof:

$$\begin{array}{rlrl}{L}_1& =\frac{N_1{\varphi }_1}{i_1}& L_2& =\frac{N_2{\varphi }_2}{i_2}\\ M& =\frac{N_2{\varphi }_{12}}{i_1}& M& =\frac{N_1{\varphi }_{21}}{i_2}\end{array}$$ $$MM=\frac{N_1{\varphi }_{21}{N}_2{\varphi }_{12}}{i_1{i}_2}$$ If $k={\varphi }_{12}/{\varphi }_1$ and $k={\varphi }_{21}/{\varphi }_2$ $$M^2=\frac{N_{1}k{\varphi }_2{N}_{2}k{\varphi }_1}{i_1{i}_2}$$ or $$M^2=k^2\frac{N_1{\varphi }_1}{i_1}\frac{N_2{\varphi }_2}{i_2}=k^2{L}_1{L}_2$$ therefore $$k=\frac{M}{\sqrt{L_1{L}_2}}=\frac{{\varphi }_{12}}{{\varphi }_1}=\frac{{\varphi }_{21}}{{\varphi }_2}$$