If f is a meromorphic function then f(z+c) is also so of the same order

Question

I am stuck with the statement given in the title. If $f$ is a meromorphic function, and $c\neq 0$, a complex constant, then $f(z+c)$ is also a meromorphic function of the same order. If $f$ is a periodic function of period $c$, then it is obvious. What happens when it is not periodic? Are my thoughts correct? Please help me proceed. The order of a meromorphic function is defined as $\sigma=\displaystyle{\limsup\limits_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r}}$, where $T(r,f)$ is Nevanlinna the characteristic function.