(n-1)!+1=n²
(n-1)!=n²-1
(n-1)!=(n-1)(n+1)
(n-1)(n-2)!=(n-1)(n+1)
(n-2)!=n+1
(n-2)!=n-2+3

Substitutions
x=n-2

x!=x+3

If x=1?
1!=1+3
1≠4 💀

If x=2?
2!=2+3
2≠5 💀

If x=3?
3!=3+3
6=6  ✅️

Therefore
x=3
n-2=3
n=5 ❤

There was no need to go through all those steps , but then again it was good to see other methods of working the problems.

It is way simpler graphically. Factorial grows faster than square, just draw graphs for n^2 and (n-1)!+1 and check that at about 7 it is already ahead. So just a few integers to check out.

The problem is that this can be quickly and easily solved by guessing. It's obvious that n is "smallish", because n! is rather close to n-squared. In other words, this problem can be solved without applying advanced math theory.

As a minor point, you don't need to show k is an integer  using k! = k+3, as k is  defined as n-2 which is obviously an integer.   You also divided by (n-1) without showing that this is not zero

In Germany we always start with an Ansatz (Educated guess). So I look at the equation and I think Let's try n=5. And 5 does the job!

If you remember factorial and square of first five natural numbers.

It will very easy find the solution.

You will not even need a pen and paper.

I have never seen a problem like this were guessing doesn't work. If I see ! I start guessing.

So this on prime newtons.
Such a eye opening example for beginners

This is one of the problems, when a short consideration of the facts is way faster than any kind of calculation:
1. n! is growing WAY FASTER than n² (That -1 and +1 doesn't change that at all)
2. IF there is a solution, it MUST be a small integer.
3. Test 1, 2, 3, 4, 5 ...
4. done.
Time used: some seconds ...

As you show this equation can be written as (n--2)!=n+1.By guess and check you get n=5.To prove that this is the single solution you graph the function (n--2)! and the function  n+1.It's obvious that the curve of the function (n--2)! and the line of the function n+1 have a single point of intersection at n=5.

First we know, that nmust be a whole number greater or equal to 1, because factorial is only defined for whole numbers greater to 0. We can tesst n=1 by dropping in n=1 in the original equation, and we see,that n=1 is not a solution.
With this knowledge we know that n is greater or equal to 2. For n greater or equal to 0, we know that (n-1)!=(n-1)*(n-2)!. Now we can do the following transformations on the equation:
(n-1)!+1=n^2
(n-1)*(n-2)!=n^2-1
(n-1)*(n-2)!=(n-1)*(n+1)
(n-1)*(n-2)!-(n-1)*(n+1)=0
(n-1)*((n-2)!-(n+1))=0
n-1=0  or  (n-2)!-n-1=0
n=1  or  (n-2)!=n+1
n=1 is impossible, since we already know,that n=1 is not a solution. So the solution of the original equationmust alsobe a solution of (n-2)!=n+1. For n>=2, factoriali growing much faster than n+1, so there can't be more than 1 solution. Now we can just test the values n=2, n=3, n=4 and n=5, if those values are solutions. The only value, that is a solution of this equation is n=5: (5-2)!=3!=6 and 5+1=6 Since we know, that there is only one solution, we have found this one solution: n=5

There are several mistakes in the video: First (n-1)! only equals (n-1)*(n-2) if n>=2,because otherwise, (n-2)! would be the factorial of a negative whole number, which is undefined. Next, you said, the factor (n-1) on both sides  will vancel, but that would only be the vase for (n-1) fifferent from 0 i.e. for n different from 1. For n=1, (n-2)! is undefined, so that (n-1)! would not be equal to (n-1)*(n-2)! in this case  ... So we have to check first, that n=1 is not a solution. So the first mistake was to use the qquation (n-1)!=(n-1)*(n-2)! without being sure, that (n-2)! is defined. The next mistake was to divide by (n-1) without checking, that (n-1) is different from 0 (saying "factor (n-1) cancels" is really dividing by (n-1)). The Substitution is ore elegant than "try and error" rgat i used in my solution, but there is a third mistake in the video: You used , that k!=k*(k-1)!, which is only the case for k>0, so you excluded another possible solution (n=2) without checking it (n=2 means k=0 and therefor k! is not equal to k*(k-1)! because (k-1)! is undefined for k=0). You shoulöd really learn to check the conditions, for which your transformations to the equation are valid.

If your intuition works fine, you may “see” n=5 immediately.
If you trust in the “unwritten law” of social media math-riddles mostly dealing with small integers, you may try and error starting with n=2 (0 and 1 are excluded from the get-go), and will also find n=5 within a couple of seconds.
Third, you can let your favorite CAS plot both sides of the equation as functions, and get the intersection point (5, 25) at your fingertips.
If all this is not satisfying 4u, then do it the hard algebraic way, or just try to follow the tricky (and incomplete) explanations of the AI-based computer voice in the video…
Good luck 😉!
🙂👻

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Super awesome video. I am fourth 😄

Very easy to see the two evident solutions : 5 and 1   ;  but when you simplify by n-1,  you remove the solution : n = 1. No place for you in Harvard (I hope)

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What's the name of that specific math content!? Series of equations?🤷😊😅

India se❤❤❤

5...... Maybe