Why is $ \operatorname{E}Y^2 =\operatorname{Var}Y +(EY)^2$ 
Why does $ \operatorname{E}Y^2 =\operatorname{Var}Y +(\operatorname{E}Y)^2$ work? Is the result true for only the binomial distribution? Does the result work for  work for $ \operatorname{E}Y^k =k\operatorname{Var}Y +(\operatorname{E}Y)^k$?
 A: By definition, for any random variable with finite first and second moments,
$$\begin{align*} \operatorname{Var}[X] &= \operatorname{E}[(X - \operatorname{E}[X])^2] \\
&= \operatorname{E}[X^2 - 2\operatorname{E}[X]X + \operatorname{E}[X]^2] \\
&= \operatorname{E}[X^2] - \operatorname{E}[2\operatorname{E}[X]X] + \operatorname{E}[\operatorname{E}[X]^2] \\
&= \operatorname{E}[X^2] - 2\operatorname{E}[X]\operatorname{E}[X] + \operatorname{E}[X]^2 \\
&= \operatorname{E}[X^2] - \operatorname{E}[X]^2. 
\end{align*}$$
The first step is by definition; the second step is the expansion of the square; the third step follows from the linearity of expectation; the fourth step follows from the fact that $\operatorname{E}[cX] = c\operatorname{E}[X]$ for any constant $c$; and the last step is algebraic simplification.
It is not required that $X$ be binomially distributed.
Is it true that $$\operatorname{E}[Y^k] \overset{?}{=} k \operatorname{Var}[Y] + \operatorname{E}[Y]^k$$ for general $k$?  No.  It isn't even true if $k \operatorname{Var}[Y]$ is replaced with $(k-1)\operatorname{Var}[Y]$ as it would need to be for $k = 1$ and $k = 2$ cases to hold.  If $k = 3$, the result does not hold.  I leave it to you as an exercise to show this.
A: Your proposed identity for $k\ne 2$ is false.
Let $\mu= \operatorname{E}(Y)$.  Then
\begin{align}
\operatorname{Var}(Y) & = \operatorname{E}((X-\mu)^2) = \operatorname{E}(X^2-2\mu X + \mu^2) = \operatorname{E}(X^2) - 2\mu\operatorname{E}(X) + \mu^2 \\[10pt]
 & = \operatorname{E}(X^2) -2\mu(\mu) + \mu^2 = \operatorname{E}(X^2) - \mu^2,
\end{align}
so
$$
\operatorname{E}(X^2) = \operatorname{Var}(X) + \mu^2.
$$
For $k\ne2$, an analogous identity says
$$
\operatorname{E}(X^k) = \text{a certain sum of cumulants of the distribution of } X.
$$
[to be continued, maybe $\ldots$ ]
A: For any discrete random variable $X$, you have that
\begin{align}
\operatorname{Var}X = \sum^n_{i=1}p_i\cdot (x_i-\mu)^2
\end{align}
where $p_i = P(X = x_i)$ and $\mu=EX$. Observe
\begin{align}
\sum^n_{i=1}p_i\cdot(x_i^2-2\mu x_i +\mu^2) =&\ \sum^n_{i=1}p_i\cdot x_i^2 -2\mu\sum^n_{i=1} x_i\cdot p_i +\mu^2\sum^n_{i=1} p_i\\
=&\ E[X^2] -2\mu^2 +\mu^2 = E[X^2]-[EX]^2
\end{align}
which means
\begin{align}
E[X^2] = \operatorname{Var} X+[EX]^2. 
\end{align}
However, you claim
\begin{align}
E[X^k] = k\operatorname{Var} X+[EX]^k
\end{align}
is not true for $k \neq 2$. 
