sorry for the fundamental question, but can someone please help me understand how this problem was simplified in these three lines? Step 1 to step 2 seems fairly obvious if I understand it correctly, though how does placing that invisible integer there help in the overall simplification? My main confusion comes from step 2 to 3. Again, sorry for the rather basic question, but clearing this up will greatly help me work through these problems in the future.
$\begin{array}{rlr}\sum_{i=1}^{k+1} i(i !) & =\sum_{i=1}^{k} i(i !)+(k+1)((k+1) !) & \\ & =[(k+1) !-1]+(k+1)((k+1) !) & \text { Since } P(k) \text { is true } \\ & =1 \cdot(k+1) !-1+(k+1)((k+1) !) & \\ & =(1+(k+1)) \cdot(k+1) !-1 & \text { Distributive property } \\ \end{array}$
I think only the plenty of expressions containing $k$ might distract you a bit.
In such a case you may replace given expressions by simple variables to see what actually happens as in your case between step 2 and 3.
For example: Just for the time being set $b=(k+1)!$ and $a=(k+1)$, then you see from $2$ to $3$ the following happens
$$b - 1 + ab = b+ab -1 = (1+a)b - 1$$