We can assume that $n\ge 3$ because if $n\leq 2$, then we are done since for every positive integer $k$, we have $\phi(k)\ge 1$. Let $k=p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ where $\alpha_i>0$ and $p_1<\cdots < p_t$. It is obvious that $p_i\ge i+1$. If $t\ge n-1$, then $$\phi{(k)}=\prod{(p_i-1)\left(p_i^{\alpha_i-1}\right)}\ge \prod{(p_i-1)}\ge t!\ge (n-1)!$$ If $t\leq n-2$, then $$\phi{(k)}=k\cdot \prod{\left(1-\frac 1{p_i}\right)}\ge k\cdot\prod_{j=1}^{t}{\left(\frac{j}{j+1}\right)}=\frac k{t+1}\ge \frac k{n-1}>(n-1)!$$

Hopefully this works. The bound seems quite weak anyway. Suppose $n! \leq k < (n+1)!$. The product of the $n$ smallest primes is strictly larger than $2\cdot 3\cdot\ldots\cdot (n+1) = (n+1)!$, so $k$ is divisible by at most $n-1$ distinct primes. Thus \[ \frac{\phi(k)}k = \prod_{p\mid k}\left(1 - \frac 1p\right) \geq \prod_{2\leq k \leq n}\left(1 - \frac 1k\right) = \frac 1n. \]It follows that $\phi(k) \geq \tfrac kn \geq (n-1)!$.

Observe that it is sufficient to show the inequality for the greatest $n$ which satisfies $k \geq n!$. Let $k=p_1^{r_1}p_2^{r_2} \cdots p_m^{r_m}$ be the prime factorization of $k$. It is sufficient to prove that $(1-\dfrac{1}{p_1})(1-\dfrac{1}{p_2}) \cdots (1-\dfrac{1}{p_m}) \geq \dfrac{1}{n}$, since it is known that $k \geq n!$. This is equivalent to showing that $$ n \geq \dfrac{p_1}{p_1-1} \dfrac{p_2}{p_2-1} \cdots \dfrac{p_m}{p_m-1} $$. For $m = 5$, since the RHS (call it $P_m$, for a number $k$ with $m$ prime factors) always decreases if we increase the primes in the factorization, it can be at most $\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{5}{4} \cdot \dfrac{7}{6} \cdot \dfrac{11}{10} = \dfrac{385}{80} < 5$. It follows that the above claim is true since $n \geq m$. We will show that $P_m < m$ for all $m \geq 5$. If we pass from $m$ to $m+1$, we have $P_{m+1} = P_m \cdot \dfrac{p_{m+1}}{p_{m+1}-1} < m \cdot \dfrac{p_{m+1}}{p_{m+1}-1} < m+1$ (the last inequality is easy to check since $p_{m+1} > m+1$). Now, in the light of the above observations, the claim becomes true. We only need to check the case where $m$ is $1$, $2$, $3$ or $4$. If $m = 4$, $P_4 = \dfrac{105}{24} < 5$ and $n$ must be at least $5$ since $2 \cdot 3 \cdot 5 \cdot 7 > 5!$, so the claim is also true for $m = 4$. The remaining cases can be verified similarly, and thus the inequality $\phi(k) \geq (n-1)!$ has been proved.

djmathman wrote: Hopefully this works. The bound seems quite weak anyway. Indeed, for $k \approx n!$ this corresponds roughly to $\varphi(k) \gg \frac{k}{\log k}$ whereas the sharp bound (which is not hard to prove) is $\varphi(k) \gg \frac{k}{\log \log k}$. Back to our problem, this would imply that for $k \approx n!$ we indeed have $\varphi(k) \gg \frac{n!}{\log n}$ instead of $(n-1)!=\frac{n!}{n}$.