2,3,5,6,7,10,11 Counting with Restrictions The sequence 2, 3, 5, 6, 7, 10, 11, $\ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes nor perfect fifth powers (in the form of $x^{5}$, where $x$ is an integer). What is the $1000^{\mathrm{th}}$ term of the sequence?
I've seen problems where I need to count with restrictions (like cubes and squares). I have never seen a problem with this degree.  Here is my thought process. 
Find intersection of squares and cubes. This is simple enough. For every 6th square, there will lie a cube. We can do this for everything else (intersection of 2 and 4, 2 and 5, 3 and 4, 3 and 5), but it will be tedious. Now, count out the numbers. This solution is very tedious...Can someone guide me through the solution?

The answer is $1041$
Indeed the set of squares, cubes and $5$th powers up to $1041$ has $41$ elements, namely 
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 144, 169, 
196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 
576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024

It is not necessary to enumerate them, actually.
There are $32$ squares, $10$ cubes less $3$ which are $6$th powers ($1;\;64,\;729$) and $4$ fifth powers less two which are also $10$th powers ($1;\;1024$). Which makes $41$
