Logarithm exponent in Chernoff bound

Question

I am applying Chernoff bound for a Poisson process with mean $\lg n$. I am putting $\delta =4$.

Hence,

$Pr(X<(1+4)\mu)< (\frac{e^\delta}{(1+4)^{(1+4)}})^\mu$ $ = (\frac{e^\delta}{5^5})^{\lg n}=\frac{1}{c^{\lg n}}$, where $c=5^5/e^4$.

Now how can I show, for some constant $k > 1$: $Pr(X<5 \lg n)<\frac{1}{n^k}$. I want to specifically know: How, $c^{\lg n} > n^k $ and the bound of $k$.

Thanks.

Answer 1

It was a silly question. $c^{\lg n} = n^{\lg c}$. I forgot this silly relation.