Solve the equation $\log(n)\cdot\log(n+1) = 1$ Problem: 

Solve $$\log(n)\cdot\log(n+1) = 1$$ for natural numbers less than $100$

Source: The actual problem was a multiple choice question, and it goes something like this,

Find the maximum* value of $n$ such that both $(n+1)^n$ and $n^{(n+1)}$ divide $100!$

Then there were a few options that I don't remember (one was 24, and one 25 as I remember, but they were two digit natural numbers for sure)
As I sat down to solve it, I applied the following approach:
My attempt:
Let, $$100! = K(n+1)^n$$ where $K$ is an arbitrary integer
And let, $$100! = C(n)^{(n+1)}$$ where $C$ is an arbitrary integer
To get rid of the 100! and solve for n exclusively, I did
$$\frac{100!}{K} = (n+1)^n$$
Taking natural logarithm (yes, $ln$ or $log_e$ NOT $log_{10}$) on both sides
$$ \log(\frac{100!}{K}) = (n)\log(n+1) $$
Upon differentiating** w.r.t $n$ we get,
$$ 0 = \frac{n}{n+1} + \log(n+1)$$
Upon solving the second equation as above we get
$$\log(n)\cdot\log(n+1) = 1$$
Now, how do I find $n$, I know plug and chug would work but I won't have a calculator on exam. All help appreciated!


*

*maximum maybe the maximum number from the given options. I'm utterly sorry for not having the options


** is this approach trivial?
Edit: Answer is $15$ and the differentiating approach is faulty
 A: Suppose I asked you you find x such that $x^2= 5$. You take the derivative of both sides, and get $2x = 0$. Does that look right to you? When you have $100!=K(n+1)^n$, $n$ is a constant unknown quantity, not a variable. Taking the derivative doesn't make sense, as there is no function to take the derivative of.
You need to analyze this with number theory tools, not calculus. For instance, when deciding whether $(n+1)^n$ divides $100!$ when $n =24$, you need to look at the prime factorization of $n+1$: $25=5^2$. So $(n+1)^n= 5^{2*24}= 5^{48}$. There are $20$ numbers $\alpha\leq 100$ that are divisible by $5$, and $1$ number $\alpha\leq 100$ divisible by $25$, so you have $20+1=21$ factors of $5$ in $100!$, which is less than the $48$ that you need. So $n$ can't be $24$. Since it's a multiple choice question, you can just test each option.
A: If you mean $\log{x}=\ln{x}=\log_ex$ then the reasoning is the same:
$f(x)=\ln{x}\ln(x+1)-1$ is an increasing function for $x>1$.
Since $f(1)<0$ and $f(10)>0$ our equation has an unique root: $2.307...$.
Thus, in natural numbers our equation has no roots.
A: $A)$ $\log (x)$ is a strictly increasing function.
$(B)$ $n= 3 \gt e$.
$z(n):=\log(n)\log(n+1)$.
1) $z(n=1) = 0.$
2)$ z(n=3) = \log(3)\log(4) \gt 1$, using $A,B$.
3)$z(n+1) \gt z(n) \gt 1,$   for $n \ge 3$, using A,B.
No positive integer solution for n=1,3,4,5,6....100.
Remains: $n=2$.
$z(n=2) = \log(2)\log(3) \gt 0$.
Cannot rule out $n=2$ as solution of
$z(n=2)=1$, without using more info.
