Prove (or disprove) for any $k \ge n!$, $\phi(k) \ge \phi(n!)$

I'm trying to solve this question: $$\text{For all } k, n \in \mathbb{N} \text{ where } k \ge n! \text{ show that:} \\~\\ \phi(k) \ge (n-1)! \\~\\ \text{Where } \phi \text{ is Euler's totient function}$$ First I proved that $$\phi(n!) \ge (n-1)!$$ so now I only have to prove $$\phi(k) \ge \phi(n!)$$ for every $$k \ge n!$$. (There are better ways to solve this question, but now I'm just more curious about the lemma I came up with)

Originally I speculated this would probably be a wrong assumption, but using code, I checked until $$n = 10!$$ and $$k = 300,000,000$$ and found no counterexamples.

In an effort to prove it, I assumed $$k = {p_1}^{\alpha_1}{p_2}^{\alpha_2}\dots{p_m}^{\alpha_m}$$ where $$p_1, p_2,\dots,p_m$$ are prime numbers. Then I proved it for all $$k$$ that contain a prime factor $$p_i$$ such that $${p_i}^{\alpha_i - 1} \gt n!$$: $$\text{Assume for some } i \le m, {p_i}^{\alpha_i - 1} \gt n! \\~\\ \phi(k) = \phi({p_1}^{\alpha_1}{p_2}^{\alpha_2}\dots{p_{i-1}}^{\alpha_{i-1}}{p_{i+1}}^{\alpha_{i+1}}\dots{p_m}^{\alpha_m}) \cdot \phi({p_i}^{\alpha_i}) \\~\\ \phi({p_i}^{\alpha_i}) = {p_i}^{\alpha_i - 1}({p_i} - 1) \gt {p_i}^{\alpha_i - 1} \gt n! \gt\phi(n!) \\~\\ \implies \phi(k) \gt \phi(n!)$$

The claim is false. For example, take $$k = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$$: the product of the first $$10$$ primes. The result is $$k = 6\,469\,693\,230$$, which is just a tiny bit bigger than $$13! = 6\,227\,020\,800$$. However:
• $$\phi(13!)$$ is $$1\,194\,393\,600$$, and
• $$\phi(k)$$ is only $$1\,021\,870\,080$$.
In general, every repeated prime factor in $$n!$$ is useless if you want to make the totient function as small as possible. The "primorial", which multiplies distinct primes together, is much more efficient (for its size) at minimizing the totient function. To find this counterexample, I just looked for a point when they were comparable - when the primorial was only a bit bigger than its adjacent factorial.
Here is another example we can understand by hand: let $$k = 8! \cdot \frac{33}{32} = 41\,580$$. Then we can clearly see that $$k > 8!$$ (we multiplied by $$\frac{33}{32} > 1$$), but I claim that $$\phi(k) < \phi(8!)$$.
To see this, we think about the totient function via the ratio $$\frac{\phi(m)}{m}$$: this is given by the product of $$\frac{p-1}{p}$$ over all distinct prime factors $$p$$ of $$m$$. In our case, $$k$$ still has all the same prime factors that $$8!$$ does (because we divided by $$32=2^5$$, but $$8!$$ is divisible by $$2^7$$) and also $$11$$. So we have $$\frac{\phi(k)}{k} = \frac{\phi(8!)}{8!} \cdot \frac{11-1}{11}.$$ Now rearrange this to solve for $$\phi(k)/\phi(8!)$$, and use the fact that $$k/8! = \frac{33}{32}$$. We get $$\frac{\phi(k)}{\phi(8!)} = \frac{k}{8!} \cdot \frac{10}{11} = \frac{33}{32} \cdot \frac{10}{11} = \frac{15}{16} < 1.$$ Therefore $$\phi(k) < \phi(8!)$$.