$f\in M_k (\Gamma_1(N)$ and $g(z)=f(dz)$, then $g\in M_k(\Gamma_1 (dN)$

Question

Let $N,d\in\mathbb{N}$ and $\Gamma_1(N)=\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}):a\equiv d\equiv 1\mod N,~c\equiv 0\mod N\}$.

I need a proof (references or your ideas) of:

Let $f$ be a modular form of weight $k$ on $\Gamma_1(N)$ and $g(z):=f(dz)$. Then $g$ is a modular form of weight $k$ on $\Gamma_1(dN)$.

There are two things to show:

1. g is holomorphic on the upper half plane
2. $g\left(\dfrac{az+b}{cz+d}\right)=(cz+d)^kg(z)$ for all $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1(dN)$
3. g is holomorphic on cusps

1 is clear. Can anybody explain or name references, why 2 holds? It might follow from $f\left(\dfrac{az+b}{cz+d}\right)=(cz+d)^kf(z)$ for all $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1(N)$, but how?

Thanks in advance.