Matrix multiplication subtraction

Let $M_i, N_i \in \mathbb{R}^{n \times n}$ for $i = 1, 2, \dots, k$ be some matrices. Then, for all $k \in \mathbb{N}$ , the following holds:

\begin{align*} &M_1 M_2 \dots M_k - N_1 N_2 \dots N_k = \\ &= (M_1 - N_1) N_2\dots N_k + M_1(M_2 - N_2)N_3\dots N_k + \dots + M_1 M_2 \dots M_{k-1}(M_k - N_k) \end{align*}

The proof is by induction on $k$ .

For $k = 1$ , the statement is trivially true:

M_1 - N_1 = M_1 - N_1

Assume the statement holds for $k = n$ :

\begin{align*} &M_1 M_2 \dots M_n - N_1 N_2 \dots N_n = \\ &= (M_1 - N_1) N_2 \dots N_n + M_1 (M_2 - N_2) N_3 \dots N_n + \dots + M_1 M_2 \dots M_{n-1} (M_n - N_n) \end{align*}

Then, for $k = n + 1$ , we have:

\begin{align*} &M_1 M_2 \dots M_{n}M_{n+1} - N_1 N_2 \dots N_{n}N_{n+1} = \\ &= \underbrace{(M_1 M_2 \dots M_n - N_1 N_2 \dots N_n)}_{\text{Inductive hypothesis}} N_{n+1} + M_1 M_2 \dots M_n(M_{n+1} - N_{n+1}) = \\ &= (M_1 - N_1) N_2 \dots N_n N_{n+1} + M_1 (M_2 - N_2) N_3 \dots N_n N_{n+1} + \\ &+\dots + M_1 M_2 \dots M_{n-1} (M_n - N_n)N_{n+1} + M_1 M_2 \dots M_n(M_{n+1} - N_{n+1}) \quad \square \end{align*}