Small exponent - rsa1

The associated script, ciphertext, and PDF of the solution can be found in this repository:

For this challenge, we can see that m^e is probably < N. This means m^e mod n is just m^e. Therefore, we can rearrange for m to find the plaintext.

m^e = c
<=> c^(1/e) = m

Now, if we convert the integer m we get back to ascii, we get the flag. These are the results when I run my python script:

Thus, this challenge is solved.