\begin{align*} &= (n+1)! − 1 + ( (n+1) · (n+1)! )\\ &= (n+1)! (1+n+1) − 1\\ &= (n+1)! (n+2) − 1\\ &= (n+2)! − 1\\ \end{align*}
I'm confused at how the first step goes to the second step. Can someone please clarify this?
\begin{align*} &= (n+1)! − 1 + ( (n+1) · (n+1)! )\\ &= (n+1)! (1+n+1) − 1\\ &= (n+1)! (n+2) − 1\\ &= (n+2)! − 1\\ \end{align*}
I'm confused at how the first step goes to the second step. Can someone please clarify this?
It's just the distributive law $\rm\ k!\ (1 + k)\ =\ k! + k\ k!\ $ for $\rm\ k = n+1.$
TIP $\ $ Often applications of simple laws are obfuscated in specific proofs. In such contexts, in order to abstract out the essential structure, it may help to substitute variables for expressions that play no role. Let's do this above. Substituting $\rm (n+1)!\ \to\ x,\ \ n+1\ \to\ y\ $ the first two steps become
$\rm x - 1 + y\ x\ = \ x\ (1 + y) - 1 $
which makes obvious the application of the distributive law.