Has this pattern in Pascal's triangle been already found?

Question

I found this pattern in Pascal's triangle and I was wondering if it has already been found before :

            1                                +(1*1⁰) =  1  = 0!
          1   1                           -(1*1¹)+(1*2¹) =  1  = 1!
        1   2   1                      +(1*1²)-(2*2²)+(1*3²) =  2  = 2!
      1   3   3   1                 -(1*1³)+(3*2³)-(3*3³)+(1*4³) =  6  = 3!
    1   4   6   4   1            +(1*1⁴)-(4*2⁴)+(6*3⁴)-(4*4⁴)+(1*5⁴) =  24  = 4!
  1   5   10  10  5   1       -(1*1⁵)+(5*2⁵)-(10*3⁵)+(10*4⁵)-(5*5⁵)+(1*6⁵) =  120 = 5!
1   6   15  20  15  6   1  +(1*1⁶)-(6*2⁶)+(15*3⁶)-(20*4⁶)+(15*5⁶)-(6*6⁶)+(1*7⁶) =  720 = 6!

Answer 1

We have

$$\sum_{j=0}^n (-1)^{n+j} {n\choose j} (j+1)^n \\ = n! [z^n] (-1)^n \sum_{j=0}^n (-1)^j {n\choose j} \exp((j+1)z) \\ = n! [z^n] (-1)^n \exp(z) (1-\exp(z))^n \\ = n! (-1)^n [z^n] \exp(z) ((-1)^n z^n + \cdots) \\ = n! (-1)^{2n} = n!.$$

Answer 2

I tried the following

The conjecture proposed is as followed

$\sum_{j=0}^{n} (-1)^{n+j}\binom{n}{j}(j+1)^{n}=n!$

Let $S=\sum_{j=0}^{n} (-1)^{n+j}\binom{n}{j}(j+1)^{n}\tag 1$

Now After expanding $(1+j)^n=\sum_{t=0}^{n}\binom{n}{t}(j)^{t}$

The sum $(1)$ becomes

$S=\sum_{j=0}^{n} [(-1)^{n+j}\binom{n}{j}\cdot \sum_{t=0}^{n}\binom{n}{t}(j)^{t}]\tag 2$

$S=(-1)^{n}\sum_{t=0}^{n}\sum_{j=0}^{n} [(-1)^{j}\binom{n}{j}\cdot\binom{n}{t}(j)^{t}]\tag 3$

$$=(-1)^{n}\sum_{t=0}^{n}\binom{n}{t}\cdot\sum_{j=0}^{n} [(-1)^{j}\binom{n}{j}\cdot (j)^{t}]\tag 4$$

Let $S_2(t)=\sum_{j=0}^{n} [(-1)^{j}\binom{n}{j}\cdot (j)^{t}]\tag 5$

Now

$S_2(t)=0$ $\forall$$t<n$, This can be Easily proved by differentiating expansion of $nx(1+x)^{n-1}$ again and again and multiply by $x$ On both sides every time you differentiate and finally putting $x=-1$ in obtained sum

Now $S_2(n)=\sum_{j=0}^{n} [(-1)^{j}\binom{n}{j}\cdot (j)^{n}]\tag 6$

Now the $S$ becomes

$$S=(-1)^{n}\binom{n}{n}\cdot\sum_{j=0}^{n} [(-1)^{j}\binom{n}{j}\cdot (j)^{n}]\tag 4$$

$$=\sum_{j=0}^{n} [(-1)^{n+j}\binom{n}{j}\cdot (j)^{n}]\tag 4$$

Now this Expression can be understood Using nice Combinatorial argument That is available on MSE by Brian M. Scott ;Here

Here is what It states

$$\sum_{j=0}^n(-1)^{n+j}j^n\binom{n}j=\sum_{k=0}^n(-1)^k(n-k)^n\binom{n}{n-k}=\sum_{k=0}^n(-1)^k(n-k)^n\binom{n}k\;.$$ Now imagine that you have $n$ toys to distribute to $n$ children, and you want each child to get a toy; clearly you can distribute the toys in $n!$ different ways. On the other hand, you can consider the number of ‘bad’ distributions, in which some child gets no toy. For $k=1,\dots,n$, $\binom{n}k(n-k)^n$ is the number of ways to choose $k$ of the children to receive no toy and to distribute the $n$ toys to the other $n-k$ children. Of course some of the $n-k$ potentially lucky children may end up with no toy; that’s why you need an inclusion-exclusion argument. It tells you that there are $$\sum_{k=1}^n(-1)^{k-1}\binom{n}k(n-k)^n$$ bad distributions, and since there are $n^n=(-1)^0\binom{n}n(n-0)^n$ distributions altogether, there must be $$n^n-\sum_{k=1}^n(-1)^{k-1}\binom{n}k(n-k)^n=\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^n$$good ones.

Hence good ones$=n!=\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^n$

Which proves the conjecture