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?
 A: Using the inclusion-exclusion principle, the number of positive integers from $1$ to $N$ that are in your sequence is $$f(N) = N - \lfloor N^{1/2} \rfloor - \lfloor N^{1/3} \rfloor - \lfloor N^{1/5} \rfloor + \lfloor N^{1/6} \rfloor + 
\lfloor N^{1/10} \rfloor + \lfloor N^{1/15}  \rfloor- \lfloor N^{1/30} \rfloor$$ 
Since most numbers are not powers, we might start by computing $f(1000) = 960$.  We are short by $40$, so next try $f(1040) = 999$.  Pretty close!  The next value is $f(1041) = 1000$.  So your answer is $1041$. 
A: The answer is 1042. 31 squares,7 unique cubes,2 unique fifth powers and 1 unique 7th power ,a total of 41 perfect powers less than equal to 1000. One more ,1024 a tenth power between 1000 and 1041. So 1041+1=1042.
A: Up to $1000$, there are $31$ squares. So the answer is not less than $1031$.
Up to $1600$, there are $40$ squares, $11$ cubes (some also squares) and $4$ fifth powers (ditto). So the answer is between $1031$ and $1000+40+11+4 = 1055$.
Up to $1055$, there are $32$ squares, $10$ cubes and $4$ fifth powers. You can check these individually for overlap.
A: Make a guess as to where the thousandth number might be near... lets say... $2000$.  Count how many squares are between $1$ and $1200$.  Count how many cubes, and fifth powers as well.  Subtract each of these amounts from $1200$, but then we accidentally subtracted too much, so add back in an amount equal to the number of numbers which are simultaneously squares and cubes or squares and fifth powers, etc... but then we added too much, so subtract those which are simultaneously squares cubes and fifth powers.  I.e. use inclusion-exclusion principle.
Letting $A$ be the set of squares less than or equal to $1200$, $B$ the set of cubes less than or equal to $1200$ and $C$ the set of fifth powers less than or equal to $1200$, we have the number of numbers which are none of these to be:
$1200-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$
Now, notice that those numbers which are simultaneously squares and cubes are exactly those numbers which are sixth powers.  Similarly, those numbers which are simultaneously squares and fifth powers are those numbers which are tenth powers, etc...
Finally, notice that the largest $n$'th power less than or equal to $1200$ is $\lfloor\sqrt[n]{1200}\rfloor^n$ and there are $\lfloor\sqrt[n]{1200}\rfloor$ $n$'th powers less than or equal to $1200$.
We get that there are $1200-34-10-4+3+2+1-1=1157$ positive numbers less than or equal to $1200$ that aren't second, third, or fifth powers of any integer.
We may then improve our guess and once close enough move forward or back enough integers, skipping those powers we wish to avoid, until we find the correct thousandth entry.
A: Let $B(n)$ be the number of integers in range $\{1,2,\dots, n\}$ that are not squares cubes or fifth powers.
We want to find the first number such that $B(n)=1000$.
A formula for $B(n)$ can be obtained with inclusion exclusion.
We get $B(n)=n- \lfloor \sqrt{n} \rfloor- \lfloor\sqrt[3]{n} \rfloor- \lfloor\sqrt[5]{n} \rfloor + \lfloor\sqrt[6]{n}\rfloor + \lfloor\sqrt[10]{n}\rfloor + \lfloor\sqrt[15]{n}\rfloor - \lfloor\sqrt[30]{n}\rfloor $
This is very close to $n$.
Proposed method:
Take $N_0=1000$ and take $N_{i+1}=N_i+(1000-B(N_i))$. Notice this guarantees $B(N_{i-1})<1000$
The method in action:
$N_0=1000\implies B(N_0)=1000-31-10-3+3+1+1-1=1000-40$
$N_1=1040\implies B(N_1)=1040-32-10-4+3+2+1-1=999$
$N_2=1041\implies B(N_1)=1041-32-10-4+3+2+1-1=1000$
we are done. $1041$ is the answer.

verification code in c++:
#include <bits/stdc++.h>
using namespace std;


int pot(int a, int p){
    int res=1;
    while(p){
        if(p%2) res*=a;
        a=a*a;
        p/=2;
    }
    return(res);
}

int ispow(int a, int p){
    for(int i=1;pot(i,p) <= a; i++){
        if(pot(i,p)== a) return(1);
    }
    return(0);
}

int main(){
    int count=0;
    for(int i=1; ;i++){
        int add=1;
        for(int j=2;j<=5;j++){
            if(ispow(i,j)){
                add=0;
                break;
            }
        }
        count +=add;
        if(count == 1000){
            printf("%d\n",i);
            return(0);
        }
    }
}

A: 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 $41$
