Logarithm exponent in Chernoff bound

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.

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