Suppose that f is entire and that |f(z)|≤|z|^N for sufficiently large z.

Suppose that f is entire and that |f(z)|≤|z|^N for sufficiently large z. Show that f must be a
polynomial of degree at least N.

Note that the (N+1)-th derivative (and higher) of an N-th degree polynomial equals 0.
Hence, it suffices to show that f^(N+1)(z) = 0 for all z.
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Next, if |f(z)| ≤ |z|^N does not hold for |z| "sufficiently small" (say |z| ≤ M),
then we can bound |f(z)| for |z| ≤ M via Maximum Modulus Principle by some constant B > 0,
and this value occurs on |z| = M.

So, we can conclude that |f(z)| ≤ A|z|^N for some constant A > 0 and for all z in C.
('A' can be computed effectively from B for 'small' |z| as needed.)
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Fix z₀ in C, and apply Generalized Cauchy's Integral Formula to f at z₀:
f^(N+1)(z₀) = ((N+1)!/(2πi)) ∫c f(z) dz/(z - z₀)^(N+2),
where C is the circle |z - z₀| = R (for some R > 0).

Taking absolute values,
|f^(N+1)(z₀)| = |((N+1)!/(2πi)) ∫c f(z) dz/(z - z₀)^(N+2)|
..................≤ ((N+1)!/(2π)) ∫c |f(z)| |dz|/|z - z₀|^(N+2)
..................≤ ((N+1)!/(2π)) ∫c A|z^N| * |dz|/|z - z₀|^(N+2), by hypothesis
..................= ((N+1)!/(2π)) ∫(t = 0 to 2π) A|z₀ + Re^(it)|^N * |iRe^(it) dt|/|Re^(it)|^(N+2),
by parameterizing C
..................= ((N+1)!/(2π)) ∫(t = 0 to 2π) A|z₀ + Re^(it)|^N * R dt / R^(N+2)
..................≤ ((N+1)!/(2π)) ∫(t = 0 to 2π) A(|z₀| + R)^N dt / R^(N+1)
..................= ((N+1)!/(2π)) * 2π * [A(|z₀| + R)^N / R^(N+1)]
..................= A(N+1)! * (|z₀| + R)^N / R^(N+1).

Letting R→∞, we find that f^(N+1)(z₀) = 0.
Since z₀ is arbitrary, we conclude that f^(n+1)(z) = 0.
==> f(z) is a polynomial of degree N.

I hope this helps!