Maximum likelihood estimate and Wald interval for binomial success probability $p$ In an interview $63\%$ of $100$ randomly chosen people gave an positive answer. $p$ is this amount from all people.
What is the maximum likelihood estimation of $p$? Calculate the wald interval of $p$!
I understand it so, that $p = 0.63 $. 
I also want to give the steps of the deviation of the formula of the likelihood. So I tried to start with this formula for the likelihood: $L(p)=\prod \limits_{i=1}^{n}p(x_i, p) = \prod \limits_{i=1}^{n}P(X_i=x_i) = \prod \limits_{i=1}^{100}P(X_i =0.63)$
$\hat{p} = \frac{1}{n} * \sum\limits_{i=1}^{n} x_i = \frac{1}{100} * 63 = 0.63$
And for the wald interval:
$\hat{p} \pm Z_{1-\frac{\alpha}{2}} * \sqrt{\frac{\hat{p} (1-\hat{p} )}{n}}$
When I use $0.63$ for $\hat{p}$ I get the interval $[0.54; 0.72]$.
So I don't really understand how to get from the formula of $L(p)$ to the value of $\hat{p}$. I'm also not sure if this is already my final result for the likelihood. I got this with the help of a book, and there seem to be some steps missing. I'm also not sure if this is in general the right way of solving it and if I used the right formula and values for the Wald interval. 
 A: You are generally on the right track, but you have not shown how to get from your first displayed equation to your
second one. For details of that, please see page 5 of these notes, one of
the many other similar derivations available online, or on this site, where the development is the same but the notation is quite different (so I am reluctant to call your Question a duplicate).
The outline of these steps is to find the logarithm $\ell(p)$ of $L(p),$ and to find
the derivative $\frac{\partial}{\partial p}\ell(p)$ of $\ell(p)$,
then finally to set the partial derivative equal to $0$ and solve for
$p$ in terms of $X$ and $n$. As shown in the first link, the solution is
$\hat p = X/n.$ (In addition, you should check to make sure that you have
found a relative maximum, not minimum or saddle-point by setting the
derivative equal to zero.)
The particular situation for your Question is shown graphically below. The curve is the likelihood
function $L(p)$ and the red line indicates that its maximum is at $p = .63.$

If $X \sim \mathsf{Binom}(n = 100, p),$ then the maximum likelihood
estimator (MLE) $\hat p$ of $p$ is $\hat p = X/n$ $= 63/100$ $= 0.63$ in your
problem, as you say.
The 95% Wald confidence interval (CI) for $p$ is of the form
$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$
where 1.96 cuts 2.5% from the upper tail of the standard normal
distribution and $\hat p$ is the MLE of $p.$
In your problem, this CI computes to $(0.535, 0.725),$ which
matches your computation.
Note: The Wald interval is derived using asymptotic arguments, and so it works best for large $n$. In particular, the Wald interval makes two approximations: (a) the normal approximation to the binomial
distribution and (b) the estimation of the standard error of $\hat p$ which
is $\text{SE} = \sqrt{\frac{p(1-p)}{n}}$ by $\widehat{\text{SE}} = \sqrt{\frac{\hat p(1-\hat p)}{n}}.$ The combination of these two approximations
causes the actual coverage probability of the Wald interval to be considerably
different from 95% (often less than 95%) for $n < 500$ or $1000.$   The Wikipedia article on 'binomial distribution' discusses other styles of
CIs for binomial $p$ that have more dependable coverage probabilities.
