Is this sum equivalence done right?

Question

Is $$\sum_{k = 0}^{\infty} \frac{t^{k}k}{k!} = \sum_{k = 1}^{\infty} \frac{t^{k}}{(k-1)!} = \sum_{k = 0}^{\infty} \frac{t^{k+1}}{(k)!} = \sum_{k = 0}^{\infty} \frac{t^{k}t}{(k)!} = e^t t$$ a valid equivalence?