Jack D'Aurizio's answer doesn't have any explicit bound for what $n$ is large enough to cut off the case work. Here's a possible way around this.
EDIT:
I managed to prove this by elementary means. Below is my old approach using properties of the $\Gamma$ function.
I'll start with the form $(n+1)((n+1)!)^{1/(n+1)}-n(n!)^{1/n}<n+1$.
Using AG, it's simple to see that $n!^{1/n}<(n+1)/2$, so we can deal with the following inequalities instead:
$$
(n+1)((n+1)!)^{1/(n+1)}-n(n!)^{1/n}<2(n!)^{1/n}\\
(n+1)((n+1)!)^{1/(n+1)}<(n+2)(n!)^{1/n}\\
(n+1)^{n+1}(n+1)n!<(n+2)^{n+1}n!(n!)^{1/n}\\
n+1<\left(1+\frac{1}{n+1}\right)^{n+1}(n!)^{1/n}\\
n+1<\left(1+\frac{1}{n}\right)^n(n!)^{1/n}
$$
In the last line I used the well-known inequality $\left(1+\frac{1}{n+1}\right)^{n+1}\geq\left(1+\frac{1}{n}\right)^n$. So now we have:
$$
\left(\frac{n^n}{(n+1)^{n-1}}\right)^n<n!
$$
We can use induction to get rid of $n!$. The inequality cleary holds for $n=2$. Then:
$$
(n+1)!=(n+1)n! >(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n
$$
We could complete the induction step if we knew that
$$
(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n\geq\left(\frac{(n+1)^{n+1}}{(n+2)^n}\right)^{n+1}\\
1\geq\left(\frac{n+1}{n+2}\frac{(n+1)^{2n}}{n^n(n+2)^n}\right)^n\\
1\geq\frac{n+1}{n+2}\frac{(n+1)^{2n}}{n^n(n+2)^n}
$$
But rearranging a bit, we find:
$$
1\geq\left(\frac{n+1}{n+2}\right)^{n+1}\left(\frac{n+1}{n}\right)^n\\
\left(1+\frac{1}{n+1}\right)^{n+1}\geq\left(1+\frac{1}{n}\right)^n
$$
This is the well-known inequality from before, so we may now conclude our induction step:
$$
(n+1)!>(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n\geq\left(\frac{(n+1)^{n+1}}{(n+2)^n}\right)^{n+1}
$$
And we are done.
OLD:
Define $f(x)=x(x!)^{1/x}$. The statement above amounts to proving $f(x+1)-f(x)<x+1$ for $x$ large enough and checking by hand for all smaller $n$. Now suppose we managed to show that beyond a certain $x$, we had $f'(x)<x$. Then by Lagrange's theorem there would be some $\xi\in(x,x+1)$ such that
$$
f(x+1)-f(x)=f'(\xi)<\xi<x+1
$$
To prove this, we'll have to make some pretty ugly bounds for $f'$. First:
$$
f'(x)=(x!)^{1/x}\left(1-\frac{1}{x}\log(x!)+\psi(x+1)\right)
$$
Here $\psi$ is the digamma function. Wikipedia has the following estimates for $x!$:
$$
\sqrt{2\pi}x^{x+1/2}e^{-x}\leq x!\leq e^{1-x}x^{x+1/2}
$$
Also, the digamma function page has the following inequality:
$$
\psi(x)<\log x-\frac{1}{2x}
$$
Putting all this together for $x>1$:
\begin{align}
f'(x)&<(e^{1-x}x^{x+1/2})^{1/x}\left(1+\log (x+1)-\frac{1}{2(x+1)}-\frac{1}{x}\left(\frac{1}{2}\log(2\pi)+\left(x+\frac{1}{2}\right)\log(x)-x\right)\right)\\
&<e^{1/x-1}x^{1+1/2x}(2+\log(x+1)-\log(x))\\
&<e^{1/x-1}x^{1+1/2x}(2+1/x)
\end{align}
Here I dropped a bunch of negative terms in the second line. I also used $\log(1+1/x)<1/x$ in the third line. Define $g(x)=e^{1/x-1}x^{1/2x}(2+1/x)$. Then we want to prove
$g(x)<1$. Apparently (I calculated this with Mathematica because I don't feel like differentiating $g$):
$$
g'(x)=-\frac{1}{2}e^{-1+1/x}x^{-3+1/2x}(1+4x+(2x+1)\log(x))
$$
But this is negative for any $x>1$. Now we might numerically calculate that $g(9)\approx0.98$ and because $g$ is decreasing, we have $g<1$ for $x\geq 9$. Now all that remains is checking for all $n<9$.
I don't like this approach because it's too numerical at the end. At the same time, $f(x+1)-f(x)$ seems like a very well-behaved function - indeed, I'm sure there exists some way of going about this more analytically.
