Let $Y_{i}$ be IID with $P(Y_{i}=1)=\frac{1}{2}$ and $P(Y_{i}=-1)=\frac{1}{2}$. If $\Lambda_{n}=(\sum_{i=1}^{n}Y_{i})^2-n$
Show that $\Lambda_{n}$ is a Martingale for the filtration $F_{n}=\sigma(Y_{1},Y_{2},...,Y_{n})$.
I think that, by definition, $\Lambda_{n}$ is $F_{n}$ measurable but I am having some trouble in showing that $E[|\Lambda_{n}|]<\infty$ and I think I'm halfway there in showing that $E[\Lambda_{n+1}|F_{n}]=\Lambda_{n}$
Any help would be useful.
\begin{align*} \Lambda_{n+1}&=Y_{n+1}^2+(\sum_{i=1}^nY_i)^2+2Y_{n+1}\sum_{i=1}^nY_i -(n+1)\\ &=\Lambda_n+Y_{n+1}^2+2Y_{n+1}\sum_{i=1}^nY_i-1 \end{align*}
Using this as well as that $(Y_i)_i$ are iid, specifically that they are independent, that $\mathbb{E}(Y_n)=0$ $\forall n$ and that $\mathbb{E}(Y_n^2)=1$:
\begin{align*} \mathbb{E}(\Lambda_{n+1}|F_n)&=\mathbb{E}(\Lambda_n|F_n)+\mathbb{E}(Y_{n+1}^2|F_n)+2\mathbb{E}(Y_{n+1}\sum_{i=1}^nY_i|F_n)-1\\ &=\Lambda_n+\underbrace{\mathbb{E}(Y_{n+1}^2)}_{=1}+2\underbrace{\mathbb{E}(Y_{n+1})}_{=0}\mathbb{E}(\sum_{i=1}^nY_i|F_n)-1\\ &=\Lambda_n \end{align*}
And for the fact that all the involved expectation values are finite: $\Lambda_n$ is upper bounded by $n^2-n$ and lower bounded by $-n$, the expectation value thus also has these upper and lower bounds and is in particular finite.