1. Interpret the expressions

We have

$$\alpha=\frac{(4!)!}{(4!)^{3!}}, \qquad \beta=\frac{(5!)!}{(5!)^{4!}}.$$

First compute the small factorials:

$$4!=24, \qquad 3!=6,$$

so

$$\alpha=\frac{24!}{24^6}.$$

Also,

$$5!=120, \qquad 4!=24,$$

so

$$\beta=\frac{120!}{120^{24}}.$$

We need to determine whether these are natural numbers.

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2. Check $\alpha$

Write

$$24!=1\cdot 2\cdot 3\cdots 24.$$

Among these factors, the number $24$ appears once. But to show divisibility by $24^6$, it is enough to group six multiples of $24$-factors from the product.

A cleaner way is to factor:

$$24!=24\cdot 23\cdot 22\cdots 1.$$

Now note that in $24!$, there are many factors of $24$ hidden in the product. Let us count prime powers.

Since

$$24=2^3\cdot 3,$$

we need at least $2^{18}\cdot 3^6$ in $24!$.

Using Legendre's formula:

• Exponent of $2$ in $24!$:

$$\left\lfloor\frac{24}{2}\right\rfloor+\left\lfloor\frac{24}{4}\right\rfloor+\left\lfloor\frac{24}{8}\right\rfloor+\left\lfloor\frac{24}{16}\right\rfloor =12+6+3+1=22.$$

• Exponent of $3$ in $24!$:

$$\left\lfloor\frac{24}{3}\right\rfloor+\left\lfloor\frac{24}{9}\right\rfloor =8+2=10.$$

Thus $24!$ contains at least

$$2^{22}3^{10},$$

which certainly contains

$$2^{18}3^6=(2^3\cdot 3)^6=24^6.$$

Hence,

$$24^6 \mid 24! \implies \alpha=\frac{24!}{24^6}\in \mathbf N.$$

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3. Check $\beta$

We have

$$\beta=\frac{120!}{120^{24}}.$$

Since

$$120=2^3\cdot 3\cdot 5,$$

we need to check whether $120!$ contains at least

$$(2^3\cdot 3\cdot 5)^{24}=2^{72}3^{24}5^{24}.$$

Again use Legendre's formula.

• Exponent of $2$ in $120!$:

$$\left\lfloor\frac{120}{2}\right\rfloor+ \left\lfloor\frac{120}{4}\right\rfloor+ \left\lfloor\frac{120}{8}\right\rfloor+ \left\lfloor\frac{120}{16}\right\rfloor+ \left\lfloor\frac{120}{32}\right\rfloor+ \left\lfloor\frac{120}{64}\right\rfloor =60+30+15+7+3+1=116.$$

• Exponent of $3$ in $120!$:

$$\left\lfloor\frac{120}{3}\right\rfloor+ \left\lfloor\frac{120}{9}\right\rfloor+ \left\lfloor\frac{120}{27}\right\rfloor+ \left\lfloor\frac{120}{81}\right\rfloor =40+13+4+1=58.$$

• Exponent of $5$ in $120!$:

$$\left\lfloor\frac{120}{5}\right\rfloor+ \left\lfloor\frac{120}{25}\right\rfloor+ \left\lfloor\frac{120}{125}\right\rfloor =24+4+0=28.$$

Thus $120!$ contains at least

$$2^{116}3^{58}5^{28},$$

which certainly contains

$$2^{72}3^{24}5^{24}=120^{24}.$$

Therefore,

$$120^{24}\mid 120! \implies \beta=\frac{120!}{120^{24}}\in \mathbf N.$$

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4. Evaluate the options

• A: $\alpha\in\mathbf N$ and $\beta\in\mathbf N$ — True
• B: $\alpha\in\mathbf N$ and $\beta\notin\mathbf N$ — False
• C: $\alpha\notin\mathbf N$ and $\beta\in\mathbf N$ — False
• D: $\alpha\notin\mathbf N$ and $\beta\notin\mathbf N$ — False

So the correct option is

$$\boxed{A}.$$

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5. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They match.