Can the encryption exponent e be greater than ϕ(N)? So I was just wondering in RSA, can the encryption exponent e be greater than ϕ(N)??

For an examples sake, lets just say N = 707, so p = 101 &amp; q = 7. So, we have ϕ(707) = 600.

Can I have e = 707? because I can calculate d = 443 by de=1 mod 600.

I have seen it almost everywhere that 1 &lt; e &lt; ϕ(N). But my question is WHY?

Why can't I have e = 707 &amp; d = 443 in the setting I described above?
From Euler's Theorem, it turns out that, for $\gcd{(a, n)} = 1$, we have:

$$\large a^e \equiv a^{e ~ \mathrm{mod} ~ \varphi{(n)}} \pmod{n}$$

So, really, having $e &gt; \varphi{(n)}$ doesn't do anything - you may as well use the reduced exponent:

$$e ~ \mathrm{mod} ~ \varphi{(n)}$$

You can absolutely use $e = n$ if you want to, it does not leak information about the private key. It's just slow. I mean use $e = n$ directly, of course - do not reduce it modulo $\varphi{(n)}$ to make it faster, since all you would get is a reduced public exponent of $p + q - 1$, and a free ticket to instant factorization of $n$.