Why is this equality involving factorials true? $$ (n +1)! -1 +(n +1)(n +1)! = (n +2)! -1 $$
Can someone explain me how in the world is this true?
:D Thanks (yes I'm trying to understand induction).
 A: Hint:
Your equation simplifies to this:
$$(n+1)!(1+(n+1))-1=(n+2)!-1\Rightarrow (n+1)!(n+2)-1=(n+2)!-1$$ 
$$\Rightarrow (n+2)!-1=(n+2)!-1$$
See if you can figure out what I did in the last step.
A: I think you can write (n+1)! as (n+1)n! so this will result in
(n+1)n! − 1 + (n+1)(n+1)n! = (n+2)(n+1)n! − 1

-1 on left side will be annihilated by -1 on right side resulting in
(n+1)n! + (n+1)(n+1)n! = (n+2)(n+1)n!

dividing all by (n+1)n! will result in
1 + (n+1) = (n+2)

taking off parenthesis 
1 + n + 1 = n + 2

so the final result will be
n+2 = n+2

A: Both of the previous answers are logically flawed, though both can be repaired. It is not enough to show that the target equality implies an identity: if it were, we could argue that $-1=1$ implies that $(-1)^2=1^2$, which is true, so $-1=1$. It’s crucial that the steps be reversible, and this point should be made explicitly.
To avoid this problem I prefer to work with one side of the desired equation at a time. In general it’s easier to simplify a complex expression, so here’d I’d try to show that the lefthand side can be simplified to yield the righthand side. The first thing to notice is the repeated factor of $(n+1)!$:
$$\begin{align*}
(n+1)!-1+(n+1)(n+1)!&=(n+1)!\cdot\color{brown}1-\color{red}1+\color{blue}{(n+1)}(n+1)!\\
&=(n+1)!\big(\color{brown}1+\color{blue}{(n+1)}\big)-\color{red}1\\
&=(n+1)!(n+2)-1\;.
\end{align*}$$
Now you just have to remember what $(n+1)!$ and $(n+2)!$ are:
$$\color{brown}{(n+1)!}(n+2)=\color{brown}{1\cdot2\cdot3\cdot\ldots\cdot n\cdot(n+1)}\cdot(n+2)=(n+2)!\;,$$
so
$$(n+1)!(n+2)-1=(n+2)!-1\;.$$
Put the two pices together, and you’re done.
