Can you correct the following demonstration of the trace of a matrix: If $A\in M_{m\times n}(\mathbb{R}), B\in M_{n\times k}(\mathbb{R})$ and $C\in M_{k\times m}(\mathbb{R})$, then $tr(ABC)=tr(BCA)=tr(CAB)$.
$A=a_{ij}$ with $1\leq i\leq m, 1\leq j\leq n$
$B=b_{qs}$ with $1\leq q\leq n, 1\leq s\leq k$
$C=c_{tv}$with $1\leq t\leq k, 1\leq v\leq m$
$AB=ab_{is}$ with $1\leq i\leq m, 1\leq s\leq k, ab_{ij}=\sum^{i}_{w=1}a_{iw}\cdot b_{ws}$
$ABC=abc_{iv}$ with $1\leq i\leq m, 1\leq v\leq m, abc_{iv}=\sum^{k}_{z=1}ab_{iz}\cdot c_{zv}$
$abc_{iv}=\sum^{k}_{z=1}\left( \sum^{j}_{w=1}a_{iw}\cdot b_{wz}\right)\cdot c_{zv}=\sum^{k}_{z=1}\sum^{j}_{w=1}a_{iw}b_{wz}c_{zv}$
$abc_{ii}=\sum^{k}_{z=1}\sum^{j}_{w=1}a_{iw}b_{wz}c_{zi}\rightarrow tr(ABC)=\sum^{m}_{f=1}abc_{ff}=\sum^{m}_{f=1}\sum^{k}_{z=1}\sum^{j}_{w=1}a_{fw}b_{wz}c_{zf}$
$BCA=(BC)A$
$BC=bc_{qv}$ with $1\leq q\leq n, 1\leq v\leq m: bc_{qv}=\sum^{k}_{z=1}b_{qz}c_{zv}$
$BCA=bca_{qj}$ with $1\leq q\leq n, 1\leq j\leq n: bca_{qj}=\sum^{m}_{f=1}bc_{qf}a_{fj}=\sum^{m}_{f=1}\left( \sum^{k}_{z=1}b_{qz}c_{zf}\right)\cdot a_{fj}=\sum^{k}_{z=1}\sum^{m}_{f=1}a_{fj}b_{qz}c_{zf}$
$bca_{jj}=\sum^{k}_{z=1}\sum^{m}_{f=1}a_{fj}b_{jz}c_{zf}\rightarrow tr(BCA)=\sum^{j}_{w=1}bca_{ww}=\sum^{j}_{w=1}\sum^{k}_{z=1}\sum^{m}_{f=1}a_{fw}b_{wz}c_{zf}$
$CAB=(CA)B$
$CA=ca_{tj}$ with $1\leq t\leq k, 1\leq j\leq n: ca_{tj}=\sum^{k}_{z=1}c_{tz}a_{zj}$
$CAB=cab_{ts}$ with $1\leq t\leq k, 1\leq s\leq k: cab_{ts}=\sum^{m}_{f=1}ca_{tf}b_{fs}=\sum^{m}_{f=1}\left( \sum^{k}_{z=1}c_{tz}a_{zf}\right)\cdot b_{fs}=\sum^{k}_{z=1}\sum^{m}_{f=1}b_{fs}c_{tz}a_{zf}=\sum^{k}_{z=1}\sum^{m}_{f=1}a_{zf}b_{fs}c_{tz}$
$cab_{ss}=\sum^{k}_{z=1}\sum^{m}_{f=1}a_{zf}b_{fs}c_{sz}\rightarrow tr(CAB)=\sum^{j}_{w=1}cab_{ww}=\sum^{j}_{w=1}\sum^{k}_{z=1}\sum^{m}_{f=1}a_{fw}b_{wz}c_{zf}$
I also wanted to ask to you: In the case $m=n=k$ it is complied with that $tr(ABC)=tr(BAC)$?