Proof that a recursively defined sequence is monotonically decreasing. I am wanting to prove that the following recursive sequence is monotonic decreasing via proof by induction.
Let $ S_1 = 1, ~ S_{n+1} = \frac{n}{n+1} (S_n)^2;~  n \geq 1. $
Here is what I have so far but I feel the proof fails at my last statement and I am unsure how to correct it.
$$ \text{Base:} ~ S_1 = 1 > \frac{1}{2} = S_2 \\ $$ $$ \text{Assumption:} ~ S_{k+1} > S_{k+2} \\ $$ $$ \text{Proof for:} ~ k+2: $$ $$ \text{ie:} ~ S_{k+1} > S_{k+2} \\ \Rightarrow S_{k+2} = \frac{k+1}{k+2} (S_{k+1})^2 \\  \Rightarrow S_{k+2} =\frac{k+1}{k+2}(\frac{k}{k+1})^2S_{k}^4 \\ \Rightarrow S_{k+2} = \frac{k^2}{(k+1)(k+2)}S_{k}^4 < \frac{k^2}{(k+1)(k+1)}S_{k}^4 = [(\frac{k}{k+1})S_k^2]^2 = S_{k+1}^2 \\ \text{Since}~  S_{k+1}^2 > S_{k+2} \Rightarrow S_{k+1} > S_{k+2} $$
Is this fine or have I messed up badly?
Any help/hints is/are appreciated. 
 A: HINT: Use the fact that $x^2<x$ whenever $0<x<1$. I’ve left the details spoiler-protected.

 Note that $0<S_2<1$. Suppose that $0<S_n<1$, where $n\ge 2$. Then $0<S_n^2<S_n<1$, so $0<S_{n+1}=\frac{n}{n+1}S_n^2<S_n^2<S_n<1$; this not only tells you that $S_{n+1}<S_n$, it also establishes that $S_{n+1}$ satisfies the condition $0<S_{n+1}<1$, allowing the induction to continue.

A: I think you have made it a little more complicated than it needs to be. The change I would make is to add a little more information to the inductive step, namely that $S_{n+1}<1$ for $n\ge1$.
Base: $S_1=1>\frac12=S_2$ is good. Note that this additionally shows that $S_2<1$.
Inductive step: Assume $S_{n+1}<S_n$ for $n\ge 1$ (note this includes the Base step). Also assume we have shown $S_{n+1}<1$ (also true for Base step). We need to show that $S_{n+2}<S_{n+1}$ and $S_{n+2}<1$.
Then $S_{n+2}=\frac{n+1}{n+2} \ S_{n+1}^2$. Now note that $\frac{n+1}{n+2}< 1$. Also, $S_{n+1}^2=S_{n+1}S_{n+1}<S_{n+1}\cdot 1 = S_{n+1}$ by the second part of the inductive assumption. Putting these together,
\begin{align*}
  S_{n+2} &=\frac{n+1}{n+2} S_{n+1}^2 \\
   &< 1\cdot S_{n+1} \\
   &= S_{n+1}.
\end{align*}
This also shows that $S_{n+2}<1$ so that we have proved both portions of the inductive step.
