I'll probably leave this standing for two days before posting a summary of Gautschi's paper.
Here is the long-overdue follow-through. I have slightly changed a few notations, but this is otherwise Gautschi's original argument.
What Gautschi actually proves in his paper is the more general inequality
$\exp((s-1)\psi(n+1))\le\frac{\Gamma(n+s)}{\Gamma(n+1)}\le n^{s-1},\; 0\le s\le1,n\in\mathbb Z^{+}\tag{1}\label{1}$
where $\psi(n)$ is the digamma function.
Gautschi considers the function
$f(s)=\frac1{1-s}\log\left(\frac{\Gamma(n+s)}{\Gamma(n+1)}\right)$
over $0\le s <1$, from which we have $f(0)=\log(1/n)$ and
$\lim_{s\to 1}f(s)=-\psi(n+1)$
via l'Hôpital. Then we have
$(1-s)f'(s)=f(s)+\psi(n+s)$
and then by letting
$\varphi(s)=(1-s)(f(s)+\psi(n+s))$
we have $\varphi(0)=\psi(n)-\log n<0$, $\varphi(1)=0$, and $\varphi'(s)=(1-s)\,\psi ^{(1)}(n+s)$ (where $\psi ^{(1)}(n)$ is the trigamma function).
Now, since $\psi ^{(1)}(n+s)=\psi ^{(1)}(s)-\sum\limits_{k=0}^{n-1}\frac1{(s+k)^2}$ is always positive, we have that $\varphi(s)<0$, from which we deduce that $f(s)$ is monotonically decreasing over $0 (i.e., $f'(s)<0$). Therefore
$-\psi(n+1)\le f(s)\le\log\frac1{n}$
which is equivalent to $\eqref{1}$. The inequality in the OP can then be deduced from the inequality $\psi(n)<\log n$.