how to solve a system of permutations 
prove nPr - nP(r-­1) = (n - r)*nP(r-­1)

I've attempted this problem in vain without any solution and I do not really understand why and I was wondering if you could correct my steps
My steps:
LHS= $\frac{n!}{(n-r)!}-\frac{n!}{(n-r+1)!}$
$\frac{n!*(n-r+1)!}{(n-r)!(n-r+1)!}-\frac{(n!)}{(n-r)!(n-r+1)}$
but then that simplifies down back to:
$\frac{n!}{(n-r)}!-\frac{n!}{(n-r+1)!}$
I thus was wondering where i was making my mistakes
 A: Think back to the definition of a permutation.
Suppose you have 5 objects $A, B, C, D,$ and $E$. Now, $^5P_3$ is how many ways you can pick $3$ objects, in which order matters. How would you compute this by hand, without using the permutation formula? For the first object, you would have $5$ choices, for the second one, you would have $4$ choices, and for the last one you would have $3$ choices, so $5\times4\times3$ would be how many ways you could pick $3$ objects.
Now, we can look back at the original question. We can simplify the expression to $^nP_r=(n-r+1) \times ^nP_{r-1}$. Looking back at what we did in the last paragraph, we can see that $^5P_2 = 5\times4$, and $^5P_3 + 5\times4\times3$. Notice a pattern? We can infer that $^5P_4 = 5\times4\times3\times2$, and this is indeed correct. Now we can see that $(n-r+1)$ is really just the next number in the pattern; in other words, we can say $^5P_3 = (5-3+1) \times (5\times4)$, which simplifies to $5\times4\times3$.
That would finish the proof.
