Can the encryption exponent e be greater than ϕ(N)?
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 $n$.