Prove elementarily that $\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}$ is strictly decreasing Prove without calculus that the sequence 
$$L_{n}=\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}, \space n\in \mathbb N$$
is strictly decreasing.
 A: My proof attempt contained a fatal mistake. It seems to me that correcting it would be as hard as proving the statement from scratch, so I've decided to give up. The following two lemmas remain valid, and although strictly weaker than the desired result, they might still be useful to someone, so I've decided to leave them here. (Thanks to mercio for pointing out the mistake and apologies to everyone for the inconvenience.)
It is convenient to define $a_n=\sqrt[n]{n!}$ for $n\in\mathbb N$. Then, $L_n=a_{n+1}-a_n$ holds for $n\in\mathbb N$. We define a new sequence by $K_n=\frac{L_n}{a_n}$. Note that $K_n=\frac{a_{n+1}}{a_n}-1$.
Lemma 1. The sequence $(a_n)_{n=1}^\infty$ is strictly increasing.
Proof. Note that $a_{n+1}^{n+1}=(n+1)a_n^n$ holds for all $n\in\mathbb N$ by definition of $(a_n)_{n=1}^\infty$. This implies that $(\frac{a_{n+1}}{a_n})^n=\frac{n+1}{a_{n+1}}$ holds for all $n\in\mathbb N$. But $a_{n+1}^{n+1}=(n+1)!<(n+1)^{n+1}$, therefore we have $a_{n+1}<n+1$ or equivalently $\frac{n+1}{a_{n+1}}>1$, proving the claim. $\square$
Lemma 2. The sequence $(K_n)_{n=1}^\infty$ is strictly decreasing.
Proof. Clearly, it suffices to prove that $(K_n+1)_{n=1}^\infty$ is strictly decreasing. By definition of $K_n$, this means that we have to show that $\frac{a_{n+2}}{a_{n+1}}<\frac{a_{n+1}}{a_n}$ holds for all $n\in\mathbb N$. This is clearly equivalent to showing $a_{n+2}a_n<a_{n+1}^2$, which is the same as showing that $a_{n+2}^{n+2}a_n^{n+2}<a_{n+1}^{2n+4}$ holds for all $n$. By definition of $(a_n)_{n=1}^\infty$ this is equivalent to $(n+2)!n!a_n^2<((n+1)!)^2a_{n+1}^2$, so we only have to show that $$\frac{a_n^2}{a_{n+1}^2}<\frac{n+1}{n+2}$$ is true for all $n$. We can easily establish this fact by induction. Clearly, it holds for $n=1$. Suppose now, it holds for some $n\in\mathbb N$. We will show that it must also hold for $n+1$. To do this, note that $$\begin{align}\Biggl(\frac{a_{n+1}^2}{a_{n+2}^2}\Biggr)^{n+2}&=\frac{((n+1)!)^2a_{n+1}^2}{((n+2)!)^2}\\
&=\frac{(n!)^2(n+1)^2a_{n+1}^2}{((n+1)!)^2(n+2)^2}\\
&=\frac{a_n^{2n}(n+1)^2a_{n+1}^2}{(a_{n+1})^{2n+2}(n+2)^2}\\
&=\Biggl(\frac{a_n^2}{a_{n+1}^2}\Biggr)^n\frac{(n+1)^2}{(n+2)^2}\overset{\text{I.H.}}{<}\Biggl(\frac{n+1}{n+2}\Biggr)^{n+2}<\Biggl(\frac{n+2}{n+3}\Biggr)^{n+2}.\end{align}$$ Here "I.H." denotes the place where we used the inductive hypothesis. Taking $(n+2)$-nd roots completes the induction. $\square$
