Solving factorial equation $ x!=\left(x-1\right)! + 96 $ I'm stuck with a problem that apparently is simple.
I need to find $x$ in following equation:
$$
x!=\left(x-1\right)! + 96
$$
How can I solve it?
After some passage I've found:
$$
\left(x-1\right)\left(x-1\right)! = 96
$$
But then I don't know how to continue...
 A: If you factorize $96=(2^5)(3^1)=(1.2.3.4)\times 4$ you can notice it is already in the form $n!\times n$ for $n=4$ and there is no other way to arrange $2^5\times 3$ to make a product of consecutive numbers appear.
Therefore using your equation we get $x-1=n\iff x=5$

Edit:

*

*For larger $n$ in $x!=(x-1)!+n$ but still reasonably low, the factorization is still a quick method, because $n$ being divisible by a factorial it will factorize easily (i.e, not a product of big primes).

For instance let $n=334764638208000=(2^{19})(3^6)(5^3)(7^2)(11)(13)$
We can start by examinining $\dfrac{n}{13!}=(2^9)(3)(5)(7)$
From there, if I try to continue the sequence $\, 1,2,\cdots,13\ $ it is not difficult to figure out that $(14\times 15\times 16)\times 16$ comes next, and the solution is $x=17$.


*

*For very large $n$ this can become tedious,but we can switch to asymptotic approximation of the inverse factorial.

See this post for instance Inverse of a factorial
$$
n\sim e\exp\left(\operatorname{W}\left(\frac1{e}\log\left(\frac{n!}{\sqrt{2\pi}}\right)\right)\right)-\frac12\tag{1}
$$
Notice also that $(x-1)!\le (x-1)(x-1)!\le x!$ therefore $n$ is squeezed between two consecutive factorial.
E.g. for $n=96$ then $\overbrace{4!}^{24} \le 96\le \overbrace{5!}^{120}$ and just need to confirm that $x=5$ is effectively solution.
For large $n$, it suffice then to calculate the above approximation and test just a few cases to find the suitable $x$ (provided $n$ fits the equation).
E.g. $(334764638208000)(!^{-1})\approx 16.978$ and you just need to check which if $x=17$ solves the equation, if not check also $16$ and $18$.
A: Write $(x-1)(x-1)!$ as $(x-1)^2(x-2)!$
There are only two ways to write $96$ in this form: $2^2 \cdot 4!$ and $4^2 \cdot 3!$ (There is also $1^2 \cdot 96$ but $96$ isn't a factorial).
Inspection shows that $96=4^2\cdot 3!$ gives the solution $x=5$.
Reducing the problem to two possibilities is hardly trial and error.
A: First, we can move all the terms containing $x$ to the left side:
$x!-(x-1)!=96$
Now, we can write them in expanded form(this is sometimes helpful with problems involving multiple factorials):
$(x*(x-1)*(x-2)*(x-3)*...)-((x-1)*(x-2)*(x-3)*(x-4)*...)=96$
Aha! We can factor out a $(x-1)*(x-2)*...$, which is equivalent to a $(x-1)!$ We get:
$x(x-1)!-1(x-1)!=96$
So,
$(x-1)(x-1)!=96$
From here, we notice two things:
a) $(x-1)!$ is bigger than $(x-1)$
b) $(x-1)!$ has to be less than $96$
This greatly limits our possible choices.
From here, we can either prime factorize $96$, or notice that $96$ is a multiple of $24$. How does that help? Well, $24=4!$, and $96=4*24$. So that means $96=4*4!$
Now, we know $96=(x-1)*(x-1)!$, so $(x-1)*(x-1)!=4*4!$, so $x-1=4$, so
$\boxed{x=5}$
