Showing this equality holds We know that $a-b\geq1$. I'm  trying to show that
$$\sum_{k=1}^{a-b}\frac{(a-b-k)!}{(a+1-k)!}=\frac{a!-b!(a-b)!}{ba!b!}.$$
So I was thinking, starting from the LHS, if we set $a-b=n$, then we have
$$\sum_{k=1}^{n}\frac{(n-k)!}{(n+b+1-k)!}.$$
We can make it simpler by setting $n-k=m-1$. This way, if I'm not mistaken, we can write
$$\sum_{m=1}^{n}\frac{(m-1)!}{(m+b)!}.$$
But now what? Were these calculations even useful? Can anyone please guide me?
 A: We seek to show that
$$\sum_{k=1}^{a-b} \frac{(a-b-k)!}{(a+1-k)!}
= \frac{1}{b} \left[\frac{1}{b!}-\frac{(a-b)!}{a!}\right]$$
Recall from MSE 4316307
the following identity which was proved there: with $1\le k\le n$
$${n\choose k}^{-1}
= k [z^n] \log\frac{1}{1-z}
(-1)^{n-k} (1-z)^{n-k}.$$
We thus have with positive integers $a,b$ where $a-b\ge 1$ that
$$\sum_{k=1}^{a-b} \frac{(a-b-k)!}{(a+1-k)!}
= \sum_{k=0}^{a-b-1} \frac{k!}{(b+1+k)!}
= \frac{1}{(b+1)!} 
\sum_{k=0}^{a-b-1} {b+1+k\choose k}^{-1}
\\ = \frac{1}{(b+1)!} +
\frac{1}{(b+1)!} 
\sum_{k=1}^{a-b-1} {b+1+k\choose k}^{-1}
\\ = \frac{1}{(b+1)!} +
\frac{1}{(b+1)!} 
\sum_{k=1}^{a-b-1} k [z^{b+1+k}] 
\log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}.$$
We may lower $k$ to zero because there is zero contribution
and get for the sum term
$$\sum_{k=0}^{a-b-1} k [z^{b+1+k}] 
\log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}
\\ = \sum_{k=b+1}^a (k-(b+1)) [z^k] 
\log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}.$$
 Two pieces 
We thus require two pieces, the first is
$$[w^m] \frac{1}{1-w} 
\sum_{k\ge 0} w^k k  [z^k] 
\log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}.$$
This is
$$[w^{m-1}] \frac{1}{1-w} 
\left.\left( \log\frac{1}{1-z} (z-1)^{b+1} \right)'\right|_{z=w}
\\ = [w^{m-1}] \frac{1}{1-w} 
\left.\left(- (z-1)^b + (b+1) \log\frac{1}{1-z} (z-1)^b\right)
\right|_{z=w}
\\ = [w^{m-1}] 
\left(((w-1)^{b-1} - (b+1) \log\frac{1}{1-w} (w-1)^{b-1}\right).$$
The second main piece is
$$- (b+1) [w^m] \frac{1}{1-w} 
\sum_{k\ge 0} w^k  [z^k] 
\log\frac{1}{1-z} (-1)^{b+1} (1-z)^{b+1}
\\ = (b+1) [w^m] \log\frac{1}{1-w} (w-1)^b.$$
 Evaluating the pieces at $m=a$ and $m=b$ 
Evaluating at $m=a$ and $m=b$ we get for the first one
$$- \frac{b+1}{a-b} {a-1\choose a-b}^{-1}$$
and the second one
$$1 - (b+1) [w^{b-1}] \log\frac{1}{1-w} (w-1)^{b-1}.$$
Evaluate the second piece again at $m=a$ and $m=b$ we find
$$\frac{b+1}{a-b} {a\choose a-b}^{-1}$$
and
$$(b+1) [w^b] \log\frac{1}{1-w} (w-1)^b.$$
We evidently require
$$(-1)^b \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{b+1}} (1-w)^b \log\frac{1}{1-w}$$
This was also evaluated at the cited link and found to be $-H_b.$
 Collecting everything 
We obtain at last for the sum component
$$- \frac{b+1}{a-b} {a-1\choose a-b}^{-1}
-1 - (b+1) H_{b-1} 
+ \frac{b+1}{a-b} {a\choose a-b}^{-1}
+ (b+1) H_b
\\ = \frac{1}{b}
- \frac{b+1}{a-b} {a\choose b}^{-1} \frac{a}{b}
+ \frac{b+1}{a-b} {a\choose b}^{-1}
= \frac{1}{b} 
+ \frac{b+1}{a-b} {a\choose b}^{-1} \frac{b-a}{b}.$$
We get for the complete sum
$$\frac{1}{(b+1)!} + \frac{1}{b\times (b+1)!} 
- \frac{(a-b)!}{b\times a!},$$
which is the claim.
A: As you mentioned, we can change the summation index with $k=a-b+1-n$ and write the sum as
$$\sum_{n=1}^{a-b}\frac{(n-1)!}{(n+b)!} $$
Then, fix $b$ and prove the claim by induction on $a$. Given the restriction $a\ge b+1$, the base case is $a=b+1$. For the inductive step, you'd have
$$\sum_{n=1}^{a-b+1}\frac{(n-1)!}{(n+b)!}=\sum_{n=1}^{a-b}\frac{(n-1)!}{(n+b)!}+\frac{(a-b)!}{(a+1)!} $$
Use the induction hypothesis and simplify. I hope you can take it from here.
