Show that $\sum_{j=1}^{n-1} M_j(M_{j+1} - M_j) = \frac{1}{2}\left( \sum_{j=1}^n X_j \right)^2 - \frac{n}{2}$ The following question is taken from Steven Shreve Stochastic Calculus for Finance Volume 1, question $2.5$

Toss a coin repeatedly. Assume the probability of head on each toss is $\frac{1}{2},$ as is the probability of tail. Let $X_j = 1$ if the $j$th toss results in a head and $X_j = -1$ if the $j$th toss results in a tail. Consider the stochastic process $M_0,M_1,...$ defined by $M_0 = 0$ and 
  $$M_n = \sum_{j=1}^n X_j, \quad n\geq 1.$$
  Define $I_0 = 0$ and 
  $$I_n = \sum_{j=1}^{n-1} M_j(M_{j+1}  - M_j),\quad n=1,2,...$$
  Show that 
  $$I_n = \frac{1}{2}M_n^2 - \frac{n}{2}.$$

My attempt:
$$\begin{align*}
\frac{1}{2}M_n^2 - \frac{n}{2} & = \frac{1}{2} \left( \sum_{j=1}^n X_j^2 + 2\sum_{i\neq j}X_i\cdot X_j \right) - \frac{n}{2} \\
& = \frac{1}{2} \left( n + 2\sum_{i\neq j}X_i\cdot X_j  \right) - \frac{n}{2} \\
& = I_n.
\end{align*}$$
Is there an easier / shorter approach to solve this question?

Note that since $X_j^2 = 1$ and $M_1^2 = X_1^2 = 1$,
$$I_n = \sum_{j=1}^{n-1} \frac{1}{2}(M_j+M_{j+1})(M_{j+1}  - M_j) - \sum_{j=1}^{n-1} \frac{1}{2}X_{j+1}(M_{j+1}  - M_j) \\=  \frac{1}{2}\sum_{j=1}^{n-1}(M_{j+1}^2- M_j^2)- \frac{1}{2}\sum_{j=1}^{n-1}X_{j+1}^2 = \frac{1}{2} M_n^2 - \frac{1}{2} M_1^2 -\frac{n-1}{2} \\ = \frac{1}{2} M_n^2 - \frac{n}{2}$$
