Why do we estimate mean using MLE when we already know that mean is average of the data? I have come across a problem in textbook to estimate mean. 
The textbook problem is as follows:

Assume that $N$ data points, $x_1$, $x_2$, . . . , $x_N$ , have been
  generated by a one-dimensional Gaussian pdf of unknown mean, but of
  known variance. Derive the ML estimate of the mean.

My question is, Why do we need to estimate mean using MLE when we already know that mean is average of the data? The solution also says that MLE estimate is the average of the data. Do I need to do all the tiring maximizing MLE steps to find out that mean is nothing but average of the data i.e. $(x_1+x_2+\cdots+x_N)/N$ ? 
 A: It is a matter of confusing vocabulary, as illustrated by those quotes, straight from google:

average
  noun: average; plural noun: averages
  
  
*
  
*a number expressing the central or typical value in a set of data, in particular the mode, median, or (most commonly) the mean, which is
  calculated by dividing the sum of the values in the set by their
  number.
  "the proportion of over-60s is above the EU average of 19 per cent"
  synonyms: mean, median, mode, midpoint, centre
  

Not the best definition, I agree! Especially when suggesting mean as a synonym. I would think average is most appropriate for datasets or samples as in $\bar{x}$ and should not be used for distributions, as $\mu$ in $\mathfrak{N}(\mu,\sigma²)$.

mean
In mathematics, mean has several different definitions depending on
  the context.
In probability and statistics, mean and expected value are used
  synonymously to refer to one measure of the central tendency either of
  a probability distribution or of the random variable characterized by
  that distribution. In the case of a discrete probability distribution
  of a random variable X, the mean is equal to the sum over every
  possible value weighted by the probability of that value; that is, it
  is computed by taking the product of each possible value x of X and
  its probability P(x), and then adding all these products together,
  giving $\mu = \sum x P(x)$.
For a data set, the terms arithmetic mean, mathematical expectation,
  and sometimes average are used synonymously to refer to a central
  value of a discrete set of numbers: specifically, the sum of the
  values divided by the number of values. The arithmetic mean of a set
  of numbers $x_1, x_2, ..., x_n$ is typically denoted by $\bar{x}$,
  pronounced "x bar". If the data set were based on a series of
  observations obtained by sampling from a statistical population, the
  arithmetic mean is termed the sample mean (denoted $\bar{x}$) to
  distinguish it from the population mean (denoted $\mu$ or $\mu_x$).

As suggested by this Wikipedia entry, mean applies to both distributions and samples or datasets. The mean of a dataset or sample is also the mean of the empirical distribution associated with this sample. The entry also exemplifies the possibility of a confusion between the terms since it gives average and expectation as synonyms.

expectation 
  noun: expectation; plural noun: expectations

  
*Mathematics:
  another term for expected value.
  

I would restrict the use of expectation to an object obtained by an integral, as in $$\mathbb{E}[X]=\int_\mathcal{X} x\text{d}P(x)$$ but the average of a sample is once again the expectation associated with the empirical distribution derived from this sample.
A: 
Why do we need to estimate mean using MLE when we already know that mean is average of the data?

The text book problem states that $x_1,x_2,\dots,x_N$ is from $$x\sim\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
They tell you that $\sigma$ is known, but $\mu$ has to be estimated.
Is it really that obvious that a good estimate $\hat\mu=\bar x$?!
Here, $\bar x=\frac{1}{N}\sum_{i=1}^Nx_i$. 
It wasn't obvious to me, and I was quite surprised to see that it is in fact MLE estimate. 
Also, consider this: what if $\mu$ was known and $\sigma$ unknown? In this case MLE estimator is $$\hat\sigma^2=\frac{1}{N}\sum_{i=1}^N(x-\bar x)^2$$
Notice, how this estimator is not the same as a sample variance estimator! Don't "we already know" that the sample variance is given by the following equation?
$$s^2=\frac{1}{N-1}\sum_{i}(x-\bar x)^2$$
A: In this case, the average of your sample happens to also be the maximum likelihood estimator. So doing all the work derive the MLE feels like an unnecessary exercise, as you get back to your intuitive estimate of the mean you would have used in the first place. Well, this wasn't "just by chance"; this was specifically chosen to show that MLE estimators often lead to intuitive estimators. 
But what if there was no intuitive estimator? For example, suppose you had a sample of iid gamma random variables and you were interested in estimating the shape and the rate parameters. Perhaps you could try to reason out an estimator from the properties you know about Gamma distributions. But what would be the best way to do it? Using some combination of the estimated mean and variance? Why not use the estimated median instead of the mean? Or the log-mean? These all could be used to create some sort of estimator, but which will be a good one?
As it turns out, MLE theory gives us a great way of succinctly getting an answer to that question: take the values of the parameters that maximize the likelihood of the observed data (which seems pretty intuitive) and use that as your estimate. In fact, we have theory that states that under certain conditions, this will be approximately the best estimator. This is a lot better than trying to figure out a unique estimator for each type of data and then stepping lots of time worrying if it's really the best choice. 
In short: while MLE doesn't provide new insight in the case of estimating the mean of normal data, it in general is a very, very useful tool. 
