How many operations at maximum needed to compute powering of a number, i.e. $N^k$? Suppose $N$ is $L$-bit long and $k$ is also $L$-bit long. 

How to show that it takes at most O$(L^2)$ operations to compute it. 

For example $N = 1010$ and $k = 10$. Then $N^k = 6^2 = 100100$. But I am not familiar with how $100100$ comes up without going back to decimal and using the fact that $6*6 = 36$. Therefore, I also find it hard to realize that it takes at most O$(L^2)$ operations to compute the powering of a number.
It depends on what you mean by operations. If multiplication of unbounded numbers is considered one operation, then repeated squaring takes at most $2L$ operations. In that case we don't care about the size of $N$.

If we only consider multiplication of $L$-bit integers as a unit operation, then multiplying $ML$-bit integers takes $O(M^2)$ operations using the high-school algorithm (in fact, much better algorithms are known), and so repeated squaring, in which the integers have size $L,2L,4L,\ldots,2^LL$, takes $O(1^2 + 2^2 + \cdots + 2^{2L}) = O(2^{2L}L)$ operations. Using fancier multiplication algorithms that take $\tilde{O}(L)$ operations (if they exist in this setting) reduces the number of operations to $\tilde{O}(2^L)$.