These are not the factorials you are looking for...

First of all, we can write the super factorial in a different way. When we multiply $n!$ by $(n-1)!$ we get: $\begin{aligned} n(n-1)(n-2)\dots 3 \cdot 2 \cdot 1 \cdot (n-1)(n-2)\dots 3 \cdot 2 \cdot 1 &= n(n-1)(n-1)(n-2)(n-2) \dots 3\cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1 \\ &= n(n-1)^2 (n-2)^2 \dots 3^2 \cdot 2^2 \cdot 1^2 \end{aligned}$ If we do the same thing when multiplying $(n-2)!$ , we start to see a pattern appearing: $\begin{aligned} n(n-1)^2 (n-2)^2 \dots 3^2 \cdot 2^2 \cdot 1^2 \cdot (n-2)(n-3)\dots 3 \cdot 2 \cdot 1 &= n(n-1)^2(n-2)^2(n-2)(n-3)^2(n-3)\dots 3^2 \cdot 3 \cdot 2^2 \cdot 2 \cdot 1^2 \cdot 1 \\ &= n(n-1)^2(n-2)^3(n-3)^3 \dots 3^3 \cdot 2^3 \cdot 1^3 \end{aligned}$ Using this pattern, we can see that $n! (n-1)! (n-2)! \dots 3! 2! 1!$ can be written as $\boxed{n(n-1)^2(n-2)^3 \dots 3^{n-2} \cdot 2^{n-1} \cdot 1^n }$ Now when we multiply this with $n^n (n-1)^{n-1} (n-2)^{n-2} \dots 3^3 2^2 1^1$ , which is the hyper factorial, we can see that it is the same as $\begin{aligned} n\$ \cdot H(n) &= n \cdot n^n \cdot (n-1)^2(n-1)^{n-1}(n-2)^3 (n-2)^{n-2} \dots 3^{n-2} \cdot 3^3 \cdot 2^{n-1} \cdot 2^2 \cdot 1^{n} \cdot 1 \\ &= n^{n+1}(n-1)^{n+1}(n-2)^{n+1} \dots 3^{n+1} \cdot 2^{n+1} \cdot 1^{n+1} \\ &= (n(n-1)(n-2)\dots 3 \cdot 2 \cdot 1)^{n+1} \\ &= (n!)^{n+\blue{1}} \end{aligned}$ We can now see that $t = \blue{1}$ and the answer is $(3(\blue{1}))! = \green{\boxed{6}}$

According to the definition, we can substitute $n \$=n! (n-1)! (n-2)! \cdots 3! 2! 1!$ , $n! (n-1)! (n-2)! \cdots 3! 2! 1! \cdot H(n)=n^{n+t} (n-1)^{n+t} (n-2)^{n+t} \cdots 3^{n+t} 2^{n+t} 1^{n+t}$ Now divide $H(n)=n^n (n-1)^{n-1} (n-2)^{n-2} \cdots 3^3 2^2 1^1$ , we can get $n! (n-1)! (n-2)! \cdots 3! 2! 1!=n^t (n-1)^{t+1} (n-2)^{t+2} \cdots 3^{t+n-3} 2^{t+n-2} 1^{t+n-1}$ When dividing both side by $n!=n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1$ , $(n-1)! (n-2)! (n-3)! \cdots 3! 2! 1!=n^{\blue{t-1}} (n-1)^t (n-2)^{t+1} \cdots 3^{t+n-4} 2^{t+n-3} 1^{t+n-2}$ Divide both side by $(n-1)! =(n-1) (n-2) (n-3) \cdots 3 \cdot 2 \cdot 1$ , $(n-2)! (n-3)! (n-4)! \cdots 3! 2! 1!=n^{\blue{t-1}} (n-1)^{\blue{t-1}} (n-2)^t \cdots 3^{t+n-5} 2^{t+n-4} 1^{t+n-3}$ When dividing $(n-2)! = (n-2) (n-3) (n-4) \cdots 3 \cdot 2 \cdot 1$ , $(n-3)! (n-4)! (n-5)! \cdots 3! 2! 1!=n^{\blue{t-1}} (n-1)^{\blue{t-1}} (n-2)^{\blue{t-1}} (n-3)^t \cdots 3^{t+n-6} 2^{t+n-5} 1^{t+n-4}$ $\cdots \cdots \cdots \cdots$ Using this pattern, we divide both side by $(n-a)!$ continuously, $n^{t-1} (n-1)^{t-1} (n-2)^{t-1} \cdots 3^{t-1} 2^{t-1} 1^{t-1} =1$ $\boxed{\therefore (n!)^{t-1}=1}$ When $a^b=1$ , there’re two cases: $a=1, b=\text{any real number; or }a=\text{any non-zero real number, } b=0$ . To let $t$ satisfy all values of $n!$ , we choose $\text{Case II}$ . So $t-1=0$ , obviously $t=1$ . The answer is $(3t)!=(3(1))!=\boxed{6}$ .

Method 1: Proof by Induction

An easy way to solve this is using proof by induction . Let $f(n) = n\$ \cdot H(n)$ . Then $f(0) = 1 = (0!)^{0+1}$ , $f(1) = 1 = (1!)^{1+1}$ , $f(2) = 8 = (2!)^{2+1}$ , $f(3) = 1296 = (3!)^{3+1}$ , .... We note that $n\$ \cdot H(n) = (n!)^{n+1}$ is true for the first few $n$ . Assuming that it is true for $n$ , then

$\begin{aligned} (n+1)\$ \cdot H(n+1) & = (n+1)! \blue{n\$} (n+1)^{n+1} \blue{H(n)} = (n+1)! (n+1)^{n+1} \blue{(n!)^{n+1}} = ((n+1)!)^{n+2} \end{aligned}$

The claim is also true for $n+1$ . Therefore $n\$ \cdot H(n) = (n!)^{n+1}$ is true for $n \ge 0$ and $(3t)! = 3! = \boxed 6$ .

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Method 2: Factorizing $H(n)$

$\begin{aligned} H(n) & = 1^1 \cdot 2^2 \cdot 3^3 \cdots (n-2)^{n-2} (n-1)^{n-1} n^n \\ & = n! \left(\cdot 2^1 \cdot 3^2 \cdot 4^3 \cdots (n-2)^{n-3} (n-1)^{n-2} n^{n-1} \right) \\ & = \frac {(n!)^2 \left(\cdot 3^1 \cdot 4^2 \cdot 5^3 \cdots (n-2)^{n-4} (n-1)^{n-3} n^{n-2} \right)}{1!} \\ & = \frac {(n!)^3 \left(\cdot 4^1 \cdot 5^2 \cdot 6^3 \cdots (n-2)^{n-5} (n-1)^{n-4} n^{n-3} \right)}{1! \cdot 2!} \\ & = \frac {(n!)^4 \left(\cdot 5^1 \cdot 6^2 \cdot 7^3 \cdots (n-2)^{n-6} (n-1)^{n-5} n^{n-4} \right)}{1! \cdot 2! \cdot 3!} \\ & = \cdots \\ & = \frac {(n!)^n}{1! \cdot 2! \cdot 3! \cdots (n-1)!} = \frac {(n!)^n}{(n-1)\$} \\ \implies n\$ H(n) & = \frac {n\$ (n!)^n}{(n-1)\$} = n! \cdot (n!)^n = (n!)^{n+1} \end{aligned}$

Therefore $(3t)! = 3! = \boxed 6$ .